hd 5053 the Sum of Cube

                                                       the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2324    Accepted Submission(s): 971

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

21 32 5

 

Sample Output

Case #1: 36Case #2: 224
直接暴力求解：
知道1^3+2^3+3^3+...+n^3=n^2*(n+1)^2/4
顺便再加一个  1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6 
后面代码就很容易写了
#include<stdio.h>int main(){int t,j=1;__int64 n,m;scanf("%d",&t);while(t--){    __int64 sum=0;    scanf("%I64d%I64d",&n,&m);    sum=m*m*(m+1)*(m+1)/4-(n-1)*(n-1)*n*n/4;    printf("Case #%d: ",j++);    printf("%I64d\n",sum);}    return 0;}