杭电Problem 5053 the sum of cube 【数学公式】

the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2430    Accepted Submission(s): 1021

Problem Description

A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range.

 

Input

The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve.
Each case of input is a pair of integer A,B(0 < A <= B <= 10000),representing the range[A,B].

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then output the answer – sum the cube of all the integers in the range.

 

Sample Input

2
1 3
2 5

 

Sample Output

Case #1: 36
Case #2: 224

 

这道题比较水，但是是用来记录一个公式的（虽然我没用）。

1^3 + 2^3 + …… n^3 = [n (n+1） / 2]^2=（1+2+……+n)^2

#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#define MAX_N   1000005
#define TY  int
#define MAX(a, b)   ((a > b)? a: b)
#define MIN(a, b)   ((a < b)? a: b)
using namespace std;

int main() {
    int t;
    __int64 n, m;
    scanf("%d", &t);
    int cnt = 0;
    while (t--) {
        __int64 ans = 0;
        scanf("%I64d%I64d", &n, &m);
        for (__int64 i = n; i <= m; i++) {
            ans += i*i*i;
        }
        printf("Case #%d: %I64d
", ++cnt, ans);
    }
    return 0;
}