10408 - Farey sequences - UVa
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题目要求:给定一个数n,求1—n之间有多少对互质的数,
phi【i】数组表示i之前有多少个和i互质的数,
a【i】表示前phi【1】+phi【2】+……+phi【i】;
a【n】数组就是1---n之间互质的数的对数。。
#include<stdio.h>#include<string.h>long long a[1000010],phi[1000010];long long n,i,j;int main(){memset(a,0,sizeof(a));for(i=1;i<=1000000;i++)phi[i]=i;for(i=2;i<=1000000;i++){if(phi[i]==i)for(j=i;j<=1000000;j+=i)//打欧拉函数的表phi[j]=phi[j]/i*(i-1);a[i]=a[i-1]+phi[i];//求前i项和}while(scanf("%lld",&n),n){printf("%lld\n",a[n]);}}
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