题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5053
the Sum of Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 140 Accepted Submission(s): 80
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
Source
2014 ACM/ICPC Asia Regional Shanghai Online
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一道送分题，不过要小心中间过程溢出。。。
//#pragma comment(linker, "/STACK:36777216")
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("in.txt","r",stdin);
#endif
int T,kase=0;
scanf("%d",&T);
while(T--){
printf("Case #%d: ",++kase);
long long A,B;
scanf("%I64d%I64d",&A,&B);
long long sum=0;
long long sum1=(A-1)*(A)/2;
sum1=sum1*sum1;
long long sum2=(B)*(B+1)/2;
sum2=sum2*sum2;
cout<<sum2-sum1<<endl;
}
return 0;
}