hdu 5053 the Sum of Cube---2014acm上海赛区网络赛

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;
}