【hdu 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): 3099    Accepted Submission(s): 1305

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

代码：

#include<cstdio>#include<cmath>#include<algorithm>typedef __int64 LL;using namespace std;LL sum[10005];int main(){int t;scanf("%d",&t);sum[1]=1;for(int i=2;i<10005;i++)    sum[i]=sum[i-1]+(LL)i*i*i;int cs=1;while(t--){int a,b;scanf("%d%d",&a,&b);printf("Case #%d: %I64d\n",cs++,sum[b]-sum[a-1]);}return 0;}

注意一点：数组开的大点    typedef  __int64 LL;

                                            %I64d