hdu 5053 the Sum of Cube---2014acm上海賽區網絡賽

題目鏈接：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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一道送分題，不過要小心中間過程溢出。。。

//#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;
}