hdoj-5053-the Sum of Cube

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

题意：算区间立方和
题解：直接暴力

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;int main(){    ll a,b,sum;    int t;    int cas = 0;    scanf("%d",&t);    while(t--)    {        sum = 0;        scanf("%I64d%I64d",&a,&b);        for(ll i = a; i <= b; i++)        {            sum+=i*i*i;        }        printf("Case #%d: %I64d\n",++cas,sum);    }    return 0;}