HDU 2114 Calculate S(n)(公式)

Posted 2020-06-13

Calculate S(n) Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9640    Accepted Submission(s): 3492     Problem Description Calculate S(n).   S(n)=13+23 +33 +......+n3 .     Input Each line will contain one integer N(1 < n < 1000000000). Process to end of file.     Output For each case, output the last four dights of S(N) in one line.     Sample Input 1 2     Sample Output 0001 0009     Author 天邪     Source HDU 2007-10 Programming Contest_WarmUp

 

 

 1 //1~N的立方和公式为S(N) = 1^3 + 2^3 + 3^3 + … + N^3 = N^2*(N+1)^2/4
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 int main()
 7 {
 8     long long n;
 9     long long sum;
10     while(cin>>n){
11         n%=10000;
12         sum=(n*n)*(n+1)*(n+1)/4;
13         sum%=10000;
14         printf("%04lld\n",sum);
15     }
16     return 0;
17 }