hdoj--5053--the Sum of Cube(水)
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the Sum of Cube
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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
Source
2014 ACM/ICPC Asia Regional Shanghai Online
没想到那麽厉害的比赛还有这种题
#include<stdio.h>int main(){int t,num=0;long long a,b;scanf("%d",&t);while(t--){long long sum=0;scanf("%lld%lld",&a,&b);for(long long i=a;i<=b;++i)sum+=i*i*i;printf("Case #%d: %lld\n",++num,sum);}return 0;}
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