HDOJ-----5053立方和公式

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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

自己暴力过的

#include<cstdio>int main(){    int a, b, c, t;    t = 1;    scanf("%d", &a);    while(a--){        scanf("%d%d", &b, &c);        printf("Case #%d: ", t++);        long long ans = 0;        for(long long i = b; i <= c; i++){            ans += i * i * i;        }        printf("%lld\n", ans);    }    return 0;}

同学跟我说用立方和公式过更简单——

1^3 + 2^3 +...+n^3 = (n * (n+1) / 2) ^ 2

#include<cstdio>  int main()  {      int u, ans = 1;      double a, b, c, c1, c2;      scanf("%d",&u);      while(u--){          scanf("%lf%lf", &a, &b);            c1 = (a * (a-1) / 2) * (a * (a-1) / 2);          c2 = (b * (b+1) / 2) * (b * (b+1) / 2);          c = c2 - c1;          printf("Case #%d: %.lf\n", ans++, c);      }      return 0;  }

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