杭电 5053 the Sum of Cube（求区间内的立方和）打表法

转载于 2016-07-20 18:05:00 发布 · 93 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/yexiaozi/p/5689268.html

本文介绍了一个高效的算法，用于计算指定区间内所有整数立方值的总和。输入为区间范围[A,B]，输出为该范围内所有整数立方值之和。通过预处理的方式，创建了一个数组来存储每个整数及其立方值累积和，从而使得每组测试数据可以在常数时间内得到结果。

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

1 3

2 5

Sample Output

Case #1: 36

Case #2: 224

 1 #include<cstdio>
 2 __int64 a[1000000]={0,1};
 3 int main()
 4 {
 5     for(__int64 i=1;i<=1000000;i++)
 6     {
 7         a[i]=a[i-1]+i*i*i;
 8     }
 9     __int64 t,k=0;;
10     scanf("%I64d",&t);
11     while(t--)
12     {
13         __int64 m,n;
14         scanf("%I64d%I64d",&m,&n);
15         printf("Case #%I64d: %I64d\n",++k,a[n]-a[m-1]);
16     }
17 }