2 条题解
-
1
思路
根据题目提示以及高斯公式可以知道： $\\ 1^3+2^3+3^3+\cdots+n^3=(n*(n+1)/2)^2$ $\\$容易找到$2$关于$9901$的逆元是$(9901+1)/2=4951$。之后编程求出即可~代码
</p>#include<bits/stdc++.h> using namespace std; long long n ; int main( ) { cin >> n; cout<<(n%9901)(n%9901)%9901(n+1)%9901*(n+1)%9901*(4951%9901)*(4951%9901)%9901; return 0; }
-
0
hetao40577 LV 8 @ 2024-4-6 15:58:05
#include <bits/stdc++.h> using namespace std; long long mod_pow(long long base, long long exp, long long mod) { long long result = 1; base = base % mod; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % mod; } exp = exp >> 1; base = (base * base) % mod; } return result; } int main() { long long n; cin >> n; long long sum = (n % 9901) * (n % 9901 + 1) / 2 % 9901; long long result = mod_pow(sum, 2, 9901); cout << result << endl; return 0; }
- 1