更新日志
[2021年9月5日] 更新了难度1和2的题解。
[2021年9月14日] 更新了难度3和4的题解,动态规划的题目借用CSDN上的一个优秀博主(Karltan)的代码,学习了。
前言
文章内容为作者个人学习心得，解题思路及参考代码不一定是最优的，如发现有不正确的地方或更优的解法，欢迎批评指正或讨论交流，联系方式可以在页面下方找到。
1.在霍格沃茨找零钱
temp = input().split()
v1, v2 = temp[0].split("."), temp[1].split(".")
for i in range(3):
v1[i] = int(v1[i])
v2[i] = int(v2[i])
s1 = (v1[0] * 17 + v1[1]) * 29 + v1[2]
s2 = (v2[0] * 17 + v2[1]) * 29 + v2[2]
c = s2 - s1
ac = abs(c)
g = ac // (17 * 29)
s = (ac - g * 17 * 29) // 29
k = ac - g * 17 * 29 - s * 29
if c < 0:
print("-{}.{}.{}".format(g, s, k))
else:
print("{}.{}.{}".format(g, s, k))
2.最简单的计算机
def A():
global M1, R1
R1 = M1
def B():
global M2, R2
R2 = M2
def C():
global M1, R3
M1 = R3
def D():
global M2, R3
M2 = R3
def E():
global R1, R2, R3
R3 = R1 + R2
def F():
global R1, R2, R3
R3 = R1 - R2
try:
while True:
M1, M2 = input().split()
M1, M2 = int(M1), int(M2)
R1, R2, R3 = 0, 0, 0
Commands = input()
for i in Commands:
eval(i)() # 调用名字i对应的函数
print(f"{M1},{M2}")
except EOFError:
pass
3.相同生日
n = int(input())
data = []
for i in range(n):
temp = input().split()
date = int(temp[1]) * 100 + int(temp[2]) # 处理日期
flag = 0
for j in range(len(data)):
if date == data[j][0]:
data[j].append(temp[0]) # 将生日相同的学号存入同一个列表
flag = 1
break
if flag == 0:
l = [date, temp[0]]
data.append(l)
data.sort(key=lambda x: x[0]) # 按日期大小排序
for i in data:
d = i[0] % 100 # 提取日
m = (i[0] - d) // 100 # 提取月
i.pop(0) # 弹出
print("{} {} {}".format(m, d, " ".join(i)))
4.日历问题
此题解来自CSDN。
#include <iostream>
#include <string>
using namespace std;
bool checkleap(int n) //这个函数检查n年是不是闰年
{
//被400整除 或 被4整除且不被100整除，则为闰年
if (n % 400 == 0 || (n % 4 == 0 && n % 100 != 0))
return true;
return false; //否则为平年
}
int Day_Of_Year(int year)
{
if (checkleap(year))
return 366; //如果是闰年，返回366天
return 365; //否则只有365天
}
int Day_Of_Month(int year, int month) //根据年数以及月数返回该月的天数
{
if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)
return 31; //1,3,5,7,8,10,12月都有31天
else if (month == 2) //如果是二月
{
if (checkleap(year))
return 29; //如果是闰年，返回29
else
return 28; //否则返回28
}
else
return 30; //其他的月份都返回30
}
string weekdays(int n) //根据总共的天数返回处在周几
{ //可以直接试出来
string ret;
if (n % 7 == 1)
ret = "Saturday";
else if (n % 7 == 2)
ret = "Sunday";
else if (n % 7 == 3)
ret = "Monday";
else if (n % 7 == 4)
ret = "Tuesday";
else if (n % 7 == 5)
ret = "Wednesday";
else if (n % 7 == 6)
ret = "Thursday";
else
ret = "Friday";
return ret;
}
void print(int year, int month, int day, int n) //打印函数
{
if (month < 10) //如果月份数小于10，那么需要输出一个前导0
{
//天数小于10同理
if (day < 10)
cout << year << "-0" << month << "-0" << day << " " << weekdays(n) << endl;
else
cout << year << "-0" << month << "-" << day << " " << weekdays(n) << endl;
}
else
{
if (day < 10)
cout << year << "-" << month << "-0" << day << " " << weekdays(n) << endl;
else
cout << year << "-" << month << "-" << day << " " << weekdays(n) << endl;
}
}
int main()
{
int n;
while (cin >> n, n != -1)
{
n++; //包括开始这一天
int sum = n, year = 2000, month = 1, day;
//sum-天数，year-年份从2000开始，month-月份，从1开始，day-日期
//for(year=2000;sum>0;year++) sum-=Day_Of_Year(year);
while (sum > 0)
sum -= Day_Of_Year(year++); //如果当前剩余天数大于0，就将总天数减去当前年份的天数
year--; //运行至此说明sum<0，多减了一年
sum += Day_Of_Year(year); //将这多减的一年加回来
//for(month=1;sum>0;month++) sum-=Day_Of_Month(year,month);
while (sum > 0)
sum -= Day_Of_Month(year, month++); //然后一样的步骤开始算月份
month--;
sum += Day_Of_Month(year, month);
day = sum; //最后剩下的就是日期了
print(year, month, day, n); //输出
}
return 0;
}
5.小希的数表
此题解来自CSDN。
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 10010;
int n;
int sum[N]; //储存一开始输入的和
int a[N]; //储存最后算出来的每个数
int st[N]; //标记每个和出现了多少次
int main()
{
while (cin >> n, n) //多case
{
int m = n * (n - 1) / 2; //和的个数
for (int i = 0; i < m; i++) //输入和
cin >> sum[i];
for (int k = 1; k <= sum[0]; k++) //假设数列的第一个数
{
a[0] = k; //假设这个数为1,2,...,sum[0]
int j = 1; //a[0]已经确定，所以j从1开始
memset(st, 0, sizeof(st)); //将存储个数的数组初始化
for (int i = 0; i < m; i++) //用st数组来标记，以供下面判断
st[sum[i]]++;
for (int i = 0; i < m; i++)
{
if (st[sum[i]]) //这步很重要，判断sum[i]是否已经由两个数组成过
{
a[j] = sum[i] - a[0]; //假设另一个数
int t;
for (t = 0; t < j; t++) //遍历已经找到的数
{
if (st[a[j] + a[t]]) //如果a[j]+a[t]可以组成sum[]数列的一个数
st[a[j] + a[t]]--; //这个数的个数就减一
else //若组不成，则证明找的这个数不对
break;
}
if (t == j) //从t找到j一直没有冲突，说明符合题意
j++;
else
break;
}
}
if (j >= n) //找到满足要求的数列
break;
}
for (int i = 0; i < n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}
6.数塔
#include <iostream>
using namespace std;
const int N = 110;
int a[N][N], b[N][N];
int incremental[N][N];
//a数组用于求出最大路径长度
//b数组用于输出路径
//incremental数组用于存储增量
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= i; j++)
{
cin >> a[i][j];
b[i][j] = a[i][j];
}
}
for (int i = n - 2; i >= 0; i--) //从倒数第二层开始
{
for (int j = 0; j <= i; j++)
{
if (a[i + 1][j] > a[i + 1][j + 1]) //若是左＞右，那么直接往左走，增量为0
{
a[i][j] += a[i + 1][j];
}
else //否则说明左＜右，往右走，增量为1
{
a[i][j] += a[i + 1][j + 1];
incremental[i][j] = 1;
}
}
}
cout << a[0][0] << endl; //顶点位置就是最大路径值
int j = 0; //从第一个开始
for (int i = 0; i < n; i++) //遍历每一层
{
cout << b[i][j] << " "; //输出路径上的数字
j += incremental[i][j]; //加上路径上的增量这样就实现了路径的输出
}
cout << endl;
return 0;
}
7.斯诺克台球
这题直接在网上找的，建议复制代码直接提交，硬写意义不大。
#include <iostream>
#include <map>
using namespace std;
struct people
{
int n;
int a[102];
int sum;
int dangan[102];
int sss;
};
map<char, int> aa;
map<char, int> bb;
map<char, int> dd;
char cc[7] = {'r', 'y', 'g', 'c', 'b', 'p', 'B'};
int main()
{
aa['r'] = 1;
aa['y'] = 2;
aa['g'] = 3;
aa['c'] = 4;
aa['b'] = 5;
aa['p'] = 6;
aa['B'] = 7;
int t;
cin >> t;
getchar();
people A, B;
A.n = t;
B.n = t;
string x, y, X;
int m = 0, k = 0, p = 0, hong = 15, huang = 1, lve = 1, cafe = 1, lan = 1, fen = 1, hei = 1;
for (p = 0; p < t; p++)
{
hong = 15;
huang = 1;
lve = 1;
cafe = 1;
lan = 1;
fen = 1;
hei = 1;
m = 0; //m表示第几个人
A.a[p] = 0;
A.sum = 0;
A.dangan[p] = 0;
A.sss = 0;
B.a[p] = 0;
B.sum = 0;
B.dangan[p] = 0;
B.sss = 0;
x = "";
y = "";
k = 0; //k表示第几轮
while (getline(cin, X) && X != "-1")
{
x = "";
y = "";
if (X.find(" ") != string::npos)
{
x = x + X.substr(0, X.find(" "));
y = y + X.substr(X.find(" ") + 1);
}
else
{
x = x + X;
y = "";
}
dd.clear();
dd['r'] = 0;
dd['y'] = 0;
dd['g'] = 0;
dd['c'] = 0;
dd['b'] = 0;
dd['p'] = 0;
dd['B'] = 0;
for (int i = 0; i < x.length(); i++)
{
dd[x[i]]++;
}
int l = 0;
for (int i = 0; i < 7; i++)
{
if (dd[cc[i]] != 0)
l++;
}
bb.clear();
bb['r'] = 0;
bb['y'] = 0;
bb['g'] = 0;
bb['c'] = 0;
bb['b'] = 0;
bb['p'] = 0;
bb['B'] = 0;
for (int i = 0; i < y.length(); i++)
{
if (y[i] > '9' && y[i] != 'r')
{
bb[y[i]]++;
}
else if (y[i] == 'r' && (y[i + 2] > '9' || i + 2 == y.length()) && (y[i + 1] >= '1' && y[i + 1] <= '9'))
{
bb[y[i]] = bb[y[i]] + (y[i + 1] - '0');
i++;
}
else if (y[i] == 'r' && (y[i + 2] >= '0' && y[i + 2] <= '9') && (y[i + 1] >= '1' && y[i + 1] <= '9'))
{
bb[y[i]] = bb[y[i]] + (y[i + 1] - '0') * 10 + (y[i + 2] - '0');
i = i + 2;
}
}
int q = 0;
for (int i = 0; i < 7; i++)
{
if (bb[cc[i]] != 0)
q++;
}
if (x == "NULL") //x为空
{
k = 0;
if (m % 2 == 0)
{
B.a[p] = B.a[p] + 4;
m++;
}
else
{
A.a[p] = A.a[p] + 4;
m++;
}
}
else
{
if (hong > 0 && k % 2 == 0 && l == 1 && x[0] == 'r' && q == 1 && y[0] == 'r') //第一轮不犯规
{
k = 1;
if (m % 2 == 0)
{
A.a[p] += bb['r'];
A.sum += bb['r'];
}
else
{
B.a[p] += bb['r'];
B.sum += bb['r'];
}
}
else if (hong >= 0 && k % 2 != 0 && x.length() == 1 && x[0] != 'r' && q == 1 && y[0] == x[0]) //第二轮不犯规
{
k = 0;
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
}
else if (k % 2 == 0 && x[0] == 'r' && hong > 0 && l == 1 && y.length() == 0) //y为空
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
k = 0;
}
else if (k % 2 != 0 && x[0] != 'r' && hong >= 0 && l == 1 && q == 0) //y为空
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
k = 0;
}
else if (k % 2 == 0 && hong <= 0)
{
if (huang == 1)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'y' && y[0] == 'y')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
huang = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'y')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
k = 0;
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
else if (huang == 0 && lve == 1)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'g' && y[0] == 'g')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
lve = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'g')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
else if (huang == 0 && lve == 0 && cafe == 1)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'c' && y[0] == 'c')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
cafe = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'c')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
else if (lan == 1 && cafe == 0 && huang == 0 && lve == 0)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'b' && y[0] == 'b')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
lan = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'b')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
k = 0;
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
else if (fen == 1 && lan == 0 && cafe == 0 && huang == 0 && lve == 0)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'p' && y[0] == 'p')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
fen = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'p')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
else if (hei == 1 && fen == 0 && lan == 0 && cafe == 0 && huang == 0 && lve == 0)
{
if (x.length() == 1 && y.length() == 1 && x[0] == 'B' && y[0] == 'B')
{
if (m % 2 == 0)
{
A.a[p] += aa[y[0]];
A.sum += aa[y[0]];
}
else
{
B.a[p] += aa[y[0]];
B.sum += aa[y[0]];
}
hei = 0;
}
else if (x.length() == 1 && y.length() == 0 && x[0] == 'B')
{
if (m % 2 == 0)
{
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
m++;
}
else
{
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
}
else //犯规
{
k = 0;
if (m % 2 == 0)
{
//if(bb['r']>7) B.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
B.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
B.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
B.a[p] += 5;
else
B.a[p] += 4;
m++;
if (A.sum >= 100)
A.dangan[p]++;
A.sum = 0;
}
else
{
//if(bb['r']>7) A.a[p]+=bb['r'];
if (bb['B'] == 1 || dd['B'] == 1)
A.a[p] += 7;
else if (bb['p'] == 1 || dd['p'] == 1)
A.a[p] += 6;
else if (bb['b'] == 1 || dd['b'] == 1)
A.a[p] += 5;
else
A.a[p] += 4;
m++;
if (B.sum >= 100)
B.dangan[p]++;
B.sum = 0;
}
}
}
hong = hong - bb['r'];
}
cout << A.a[p] << ":" << B.a[p] << endl;
if (A.sum >= 100)
A.dangan[p]++;
if (B.sum >= 100)
B.dangan[p]++;
}
for (int i = 0; i < t; i++)
{
if (A.dangan[i] != 0)
A.sss++;
if (B.dangan[i] != 0)
B.sss++;
}
cout << A.sss << ":" << B.sss;
return 0;
}
8.最少钱币数
此题解来自CSDN。
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
int m, n;
while (cin >> m, m)
{
int f[m + 1]; //凑i元需要的硬币数的最小值
cin >> n;
int w[n];
for (int i = 0; i < n; i++)
cin >> w[i];
for (int i = 0; i <= m; i++)
f[i] = 999999;
f[0] = 0;
for (int i = 1; i <= m; i++) //遍历所有的钱数
for (int j = 0; j < n; j++) //遍历所有的硬币
if (i >= w[j]) //如果钱数大于硬币面值
f[i] = min(f[i], f[i - w[j]] + 1);
if (f[m] == 999999)
cout << "Impossible" << endl;
else
cout << f[m] << endl;
}
return 0;
}9.相等的多项式
此题解来自CSDN。
#include <iostream>
using namespace std;
const int N = 25;
int n;
int a[N], b[N], aex[N];
//a数组存每个式子的常数
//b数组存的是输入的最后的结果（下标对应次数项的系数）
//aex数组存的是最后算出来的结果（下标对应次数项的系数）
int main()
{
while (cin >> n, n)
{
for (int i = 1; i <= n; i++) //下标从1开始，因为后续有下标-1的操作
cin >> a[i];
for (int i = n - 1; i >= 0; i--)
cin >> b[i];
b[n] = 1; //最后一项的次数是1
aex[0] = a[1]; //次数为0的项的系数是第一项的常数（假设只有一个式子）
aex[1] = 1; //次数为1的项的系数是1（假设只有一个式子）
for (int i = 2; i <= n; i++)
{
aex[i] = 1; //每次将最高次数项的系数置为1
for (int j = i - 1; j > 0; j--) //遍历所有已计算的项
aex[j] = aex[j] * a[i] + aex[j - 1]; //次数为j的项 = 次数为j的项*另一项的常数 + 次数为j-1的项*1
aex[0] *= a[i]; //最后的常数项也乘上这个常数
}
int sum = 0;
for (int i = 0; i < n; i++)
if (aex[i] == b[i]) //如果对应次数项的系数相同
sum++;
if (sum == n)
cout << 'Y' << endl; //若有n个都相同，说明整个都相同
else
cout << 'N' << endl;
}
return 0;
}
10.选美比赛
n = int(input())
scores = input().split()
l = list(set(scores)) # 去除重复的分数
l.sort(key=lambda x: int(x), reverse=True) # 排名
temp = []
for i in scores:
temp.append(str(l.index(i) + 1)) # 找到某个分数对应的排名
print("{}".format(" ".join(temp)))
11.蛇行矩阵
n = int(input())
head = 1
matrix = []
for i in range(n): # 第一列的规律
head += i
matrix.append([head])
for i in range(n): # 行的规律
foot = 2 + i
while foot < n + 1:
temp = matrix[i][-1] + foot
matrix[i].append(temp)
foot += 1
for i in matrix:
for j in range(len(i)):
i[j] = str(i[j])
print(" ".join(i))
12.疫情期间
此题解来自CSDN。
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main()
{
int n;
cin >> n;
int f[n + 1][3]; //f[i][j]表示到第i天为止且第i天做了j的最小休息天数
for (int i = 0; i <= n; i++)
{
for (int j = 0; j < 3; j++)
{
f[i][j] = 999;
}
}
f[0][0] = f[0][1] = f[0][2] = 0; //如果只有0天，不管怎么样都没得休息
int x;
for (int i = 1; i <= n; i++) //由于有下标-1的操作，所以我们从1开始
{
cin >> x;
f[i][0] = min(f[i - 1][0], min(f[i - 1][1], f[i - 1][2])) + 1; //今天休息的话，昨天干什么都可以，这种情况必定存在，不需要判断
if (x == 1 || x == 3) //如果今天可以编程
f[i][1] = min(f[i - 1][0], f[i - 1][2]); //昨天可以休息和健身
if (x == 2 || x == 3) //如果今天可以健身
f[i][2] = min(f[i - 1][0], f[i - 1][1]); //昨天可以休息和编程
}
cout << min(f[n][0], min(f[n][1], f[n][2])) << endl; //第n天可以有三种情况，输出其中的最小值
return 0;
}13.7，还是7
n = int(input())
for i in range(7, n + 1):
if i % 7 == 0 or '7' in str(i):
print(i)
14.组个最小数
l = input().split()
l2 = [] # 存非0的数
zero = 0 # 0的个数
for i in range(len(l)):
l[i] = int(l[i])
if l[i] != 0:
l2.append(l[i])
else:
zero += 1
m = min(l2) # 找出最小的数当首位
l2.pop(l2.index(m)) # 弹出
l2.sort() # 剩下的数从小到大排序
num = str(m) + '0' * zero # 首位与n个0拼接
for i in l2:
num += str(i)
print(num)
15.字频统计
while True:
n = int(input())
if n == 0:
break
s = input()
t1 = s.count("ab")
t2 = s.count("ba")
print(t1 - t2)
16.逆序数
#include <iostream>
using namespace std;
int main()
{
int n;
int sum;
while (cin >> n, n)
{
sum = 0;
int a[n];
for (int i = 0; i < n; i++)
cin >> a[i];
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (a[i] > a[j])
sum++;
cout << sum << endl;
}
return 0;
}17.最小钱币数
n = int(input())
a = n // 100
b = (n - a * 100) // 20
c = (n - a * 100 - b * 20) // 10
d = (n - a * 100 - b * 20 - c * 10) // 5
e = n - a * 100 - b * 20 - c * 10 - d * 5
print(a + b + c + d + e)
18.身份证校验
w = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2]
c = ['1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2']
while True:
n = input()
if n == "-1":
break
s = 0
for i in range(17):
s += int(n[i]) * w[i]
y = s % 11
k = c[y]
if n[-1] == k:
print(1)
else:
print(0)
19.最长连续递增子序列
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int num[n];
int max = 0;
for (int i = 0; i < n; i++)
cin >> num[i];
for (int i = 0; i < n; i++)
{
int m = 1;
for (int j = i + 1; j < n; j++)
{
if (num[j] > num[j - 1])
m++;
else
break;
}
if (max < m)
max = m;
}
cout << max << endl;
return 0;
}
20.恺撒Caesar密码
def Caesar_Decrypt(Text, Move=5):
UpperCaseWord = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
]
LowerCaseWord = [
'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'
]
DecryptedText = ""
for i in Text:
if i in UpperCaseWord:
DecryptedText = DecryptedText + chr((ord(i) - ord('A') - int(Move)) % 26 + ord('A'))
elif i in LowerCaseWord:
DecryptedText = DecryptedText + chr((ord(i) - ord('a') - int(Move)) % 26 + ord('a'))
else:
DecryptedText = DecryptedText + i
return DecryptedText
while True:
start = input()
if start == "ENDOFINPUT":
break
text = input()
end = input()
print(Caesar_Decrypt(text))
21.回文串
Python的代码是对的，但是交到服务器上会出错，无法输出，用C++的代码可以正常运行(在CSDN找的)。
def isneed(s):
if s == s[::-1]:
return True
else:
return False
while True:
n = input().strip()
if n == "0":
break
s = input()
if isneed(s):
print("0")
else:
for i in range(1, len(s)):
temp = s + s[:i][::-1]
if isneed(temp):
print(str(i))
break
22 · 2022-08-22 17:15
21题是不是疏忽了忘记放c++题解了
B1ue1nWh1te · 2022-08-23 14:55 作者
是的，已经找不到了。