周赛 题 1 light oj 1005【2015/10/24】
来源:互联网 发布:wampserver mysql密码 编辑:程序博客网 时间:2024/06/16 17:27
Description
A rook is a piece used in the game of chess which is played on a board of square grids. A rook can only move vertically or horizontally from its current position and two rooks attack each other if one is on the path of the other. In the following figure, the dark squares represent the reachable locations for rook R1 from its current position. The figure also shows that the rookR1 andR2 are in attacking positions whereR1 andR3 are not. R2 and R3 are also in non-attacking positions.
Now, given two numbers n and k, your job is to determine the number of ways one can putk rooks on ann x n chessboard so that no two of them are in attacking positions.
Input
Input starts with an integer T (≤ 350), denoting the number of test cases.
Each case contains two integers n (1 ≤ n ≤ 30) and k (0 ≤ k ≤ n2).
Output
For each case, print the case number and total number of ways one can put the given number of rooks on a chessboard of the given size so that no two of them are in attacking positions. You may safely assume that this number will be less than1017.
Sample Input
8
1 1
2 1
3 1
4 1
4 2
4 3
4 4
4 5
Sample Output
Case 1: 1
Case 2: 4
Case 3: 9
Case 4: 16
Case 5: 72
Case 6: 96
Case 7: 24
Case 8: 0
分析:
这道题关键是找规律,规律sum=Cn/m* An/m.
代码:
#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;int main(){int t;long long n,m;int cnt=0;long long C,A;long long sum;scanf("%d",&t);while(t--){C=1,A=1;sum=0;long long i,j;scanf("%lld%lld",&n,&m);printf("Case %d: ",++cnt);if(m==1&&n>=m){printf("%lld\n",n*n);continue; }if(m>n){printf("0\n");continue;}else{for(i=n;i>n-m;i--){A*=i;}for(j=m;j>=1;j--){C*=j;}sum=A*(A/C);printf("%lld\n",sum); }}return 0;}
- 周赛 题 1 light oj 1005【2015/10/24】
- Light Oj 1005
- Light OJ 1005 Rooks
- 【瞎搞】 Light OJ 1005 - Rooks
- light oj
- light oj
- Light OJ
- Light OJ
- light oj 1005 - Rooks(组合数学)
- Light OJ 1005 - Rooks 数学题解
- light oj 1005 - Rooks (组合数学)
- light oj 1005 Rooks(组合数)
- Light OJ 1000
- Light OJ 1001
- Light OJ 1008
- Light OJ 1022
- Light OJ 1015
- Light OJ 1042
- mysql 学习记录(十六)--优化常用sql
- 第六周-数制转换
- JAVA多线程实现的三种方式
- 背包之01,完全,多重模板
- C++项目中的extern "C" {}
- 周赛 题 1 light oj 1005【2015/10/24】
- POJ 1325 Machine Schedule(最大匹配数=最小点覆盖)
- css min-width最小宽度与max-width最大宽度教程
- 数据库五个例子总结
- 使用延迟加载以及避免代码重复
- CSS: body{font-size: 62.5%;}为什么???
- 顺序表
- 程序员常用英语积累---持续更新
- git常用命令-Git学习笔记