the Sum of Cube<hdoj5053>
来源:互联网 发布:java怎么输入数组 编辑:程序博客网 时间:2024/05/16 08:51
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].
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
#include<cstdio>long long lifang[10001];void dabiao(){for(long long i=1;i<10001;i++){ lifang[i]=i*i*i; }}int main(){dabiao(); int t; int cut=0; scanf("%d",&t); while(t--) { long long n,m; scanf("%lld%lld",&n,&m); long long sum=0; for(long long i=n;i<=m;i++) { sum+=lifang[i];}printf("Case #%d: %lld\n",++cut,sum);}return 0;}
0 0
- the Sum of Cube<hdoj5053>
- the Sum of Cube
- The Sum of Cube
- the Sum of Cube
- the Sum of Cube
- the Sum of Cube
- I - the Sum of Cube
- HDU-#5053 the Sum of Cube
- 【水题】HDU 5053 the Sum of Cube
- HDU - 5053 the Sum of Cube
- HDU 5053 the Sum of Cube
- HDOJ 5053 the Sum of Cube
- HDU--5053 the Sum of Cube
- hdu 5053 the Sum of Cube 水题
- HDOJ-5053(the Sum of Cube)
- the Sum of Cube 就是立方和
- HDU-5053 the Sum of Cube
- 【杭电】[5053]the Sum of Cube
- git 切换到之前的版本上
- ListView同时响应itemclick事件和item控件的onClickListener时间的解决办法
- docker – 你应该知道的10件事
- 求单表序数字段中不连续的最小值
- 编码不统一问题
- the Sum of Cube<hdoj5053>
- HDU 1711 kmp入门
- Android应用界面开发04
- 朴树贝叶斯法
- 最近用到的一些sql命令(持续更新)
- Jquery中用得比较少的核心函数
- 宏碁TravelMate b115商务本一键U盘装win10教程
- SDUT 2088 refresh的停车场
- 破碎的美丽