poj 3100 (zoj 2818)||ZOJ 2829 ||ZOJ 1938 (poj 2249)

水题三题:

1.给你B和N,求个整数A使得A^n最接近B

2. 输出第N个能被3或者5整除的数

3.给你整数n和k,让你求组合数c(n,k)


1.poj 3100 (zoj 2818) Root of the Problem:

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2818

http://poj.org/problem?id=3100

#include<cstdio>
#include<cmath>
int main()
{
	int b,n;
	while(~scanf("%d%d",&b,&n),b||n)
	{
		int a=pow(b,1.0/n);
		int ans=a;
		if(b-pow(ans,n) > pow(ans+1,n)-b)
			ans++;
		printf("%d\n",ans);
	}
	return 0;
}

2.ZOJ 2829 Beautiful Number

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1829

#include<cstdio>
const int MAXN=300000;
int a[MAXN],len=1;
int main()
{
	for(int i=3;i<MAXN;i++)
		if(i % 3==0 || i %5==0)
			a[len++]=i;
	int n;
	while(~scanf("%d",&n))
	{
		printf("%d\n",a[n]);
	}
	return 0;
}


3.ZOJ 1938 Binomial Showdown (poj 2249)

http://poj.org/problem?id=2249

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=938

#include<cstdio>
typedef long long LL;
LL ans,n,k;
int main()
{
	while(scanf("%lld%lld",&n,&k),n)
	{
		ans=1;
		if(k == 0)
		{
			printf("1\n");
			continue;
		}
		if(k > n-k )k=n-k;
		for(int i=1;i<=k;i++)
		{
			ans=ans*(n-i+1)/i;
		}
		printf("%lld\n",ans);
	}
	return 0;
}



你可能感兴趣的:(编程,ACM,ZOJ,poj)