HDU - 5053 the Sum of Cube

HDU - 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

Source 2014 ACM/ICPC Asia Regional Shanghai Online
題意：不算難的題

#include 
#include 
#include 
#include 
typedef __int64 ll;
using namespace std;

int main() {
    int t, a, b, cas =1 ;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &a, &b);    
        ll sum = 0;
        for (int i = a; i <= b; i++)
            sum += (ll)i * i * i;
        printf("Case #%d: ", cas++);
        printf("%I64d\n", sum);
    }
    return 0;
}