SPOJ Problem 11:Factoral

这是一道水题，求某数阶乘结尾0的个数。

啥都不说了，上代码

#include<cstdio>
int n,x,y;
int main(){
    for (scanf("%d",&n);n--;){
        for (scanf("%d",&x),y=0;x;x/=5)y+=x/5;
        printf("%d\n",y);
    }
}
时间： 2024-12-20 06:06:10

SPOJ Problem 11:Factoral的相关文章

素数筛法--SPOJ Problem 2 Prime Generator

质数(prime number)又称素数,除了1和它本身外,不能整除以其他自然数,换句话说就是该数除了1和它本身以外不再有其他的因数:否则称为合数.最小的质数是2. 要判断一个整数N是不是质数很简单,看它是否能被2到sqrt(N)之间的整数整除即可. def isPrime(n): if n%2==0: return False for i in xrange(3,int(math.sqrt(n)+1),2): if n%i==0: return False return True 不过要找出1

SPOJ Problem 7974:What&#39;s Next

求数列下一项,啥都不说了,贡献了N多WA... #include<cstdio> int main(){ int a,b,c; while(scanf("%d%d%d",&a,&b,&c)&&(a||b||c)){ if (b*2==c+a&&b!=a) printf("AP %d\n",2*c-b); else printf("GP %d\n",c*c/b); } return

SPOJ Problem:A Game with Numbers

突然就看到SPOJ升级了,让我好不适应.. 这一一道博弈论的题目,可以先前面暴力一下,易得小于十的为必胜态,十的时候必败,然后11到19又必胜,而且发现只要各位为零且这个数不为零就必败. 再依次验证上百,上千的数.. #include<cstdio> #include<cstring> int n; int main(){ scanf("%d",&n); if (n%10==0&&n!=0)printf("2\n");

SPOJ Problem 2: Prime Generator

嗯..在SPOJ上刷的第二题. 一开始不知道哪错了,后来发现i出现了两遍.. 因为m<10^9,所以用素数筛筛32000以内的数,开一个4000的数组存储就行.然后再从n开始用素数筛,总之效率还行. 代码如下: //0.01s 3.2M #include<cstdio> #include<cstring> #include<cmath> int n,i,j,t,m,k,tot; int a[100005],b[4000]; int prime[40000]; in

SPOJ Problem 9948:Will it ever stop

如题..http://www.spoj.com/problems/WILLITST/ #include<cstdio> long long n; int main(){ scanf("%lld",&n); while(n>1){ if (n==3||n==6){printf("NIE\n");return 0;} if (n&1)n=(n+1)/2*3;else n>>=1; } printf("TAK\n&q

SPOJ Problem 302:Count on Cantor

题目不复述了,自己看吧.http://www.spoj.com/problems/CANTON/一如既往的暴力.. #include<cstdio> int n,i,j,s,ans; int main(){ scanf("%d",&n); while(n--){ scanf("%d",&s); printf("TERM %d IS ",s); for (i=1;s>i;i++){ s-=i; } if (i&am

SPOJ Problem 379:Ambiguous Permutations

http://www.spoj.com/problems/PERMUT2/ 如题.. #include<cstdio> #include<cstring> int n,i,j; int a[100005]; int main(){ while(scanf("%d",&n)&&n){ for (i=1;i<=n;i++){ scanf("%d",&a[i]); } for (i=1;i<=n;i++)

SPOJ Problem 2727:Army Strength

题目在此  http://www.spoj.com/problems/ARMY/ 总体来说就是求最大值.. #include<cstdio> #include<cstring> int x,n,t,i,a,b; int max,pos; int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&a,&b); max=0; for (i=1;i<=a;i++){ s

Project Euler:Problem 11 Largest product in a grid

In the 20×20 grid below, four numbers along a diagonal line have been marked in red. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30