2 条题解
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0
正解:
#include<bits/stdc++.h> using namespace std; int fastPowMod(int a, int b, int m) { int result = 1; a = a % m; while (b > 0) { if (b % 2 == 1) { result = (result * a) % m; } a = (a * a) % m; b = b / 2; } return result; } int main() { int a, b; cin >> a >> b; int mod = fastPowMod(a, b, 7); string weekdays[7] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}; cout << weekdays[mod] << endl; return 0; }
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0
#include
#include
using namespace std;
// 快速幂计算 a^b mod m
int fastPowMod(int a, int b, int m) { int result = 1;
a = a % m; // 先取模,防止a过大 while (b > 0) { if (b % 2 == 1) { result = (result * a) % m; } a = (a * a) % m; b = b / 2; } return result;}
int main()
{
int a, b; cin >> a >> b; // 计算 a^b mod 7 int mod = fastPowMod(a, b, 7); string weekdays[7] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}; cout << weekdays[mod] << endl; return 0;}
- 1
信息
- ID
- 87
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 4
- 标签
- 递交数
- 60
- 已通过
- 26
- 上传者