标签：gif 技术 spl head 次数 lines hint div inpu
It‘s Bessie‘s birthday and time for party games! Bessie has instructed the N (1 < N < 100,000) cows conveniently numbered 1..N to sit in a circle (so that cow i [except at the ends] sits next to cows i-1 and i+1; cow N sits next to cow 1). Meanwhile, Farmer John fills a barrel with one billion slips of paper, each containing some integer in the range 1..1,000,000.
Each cow i then draws a number Ai (1 ≤ Ai ≤ 1,000,000) (which is not necessarily unique, of course) from the giant barrel. Taking turns, each cow i then takes a walk around the circle and pats the heads of all other cows j such that her number Ai is exactly divisible by cow j‘s number Aj; she then sits again back in her original position.
The cows would like you to help them determine, for each cow, the number of other cows she should pat.
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single integer: Ai
* Lines 1..N: On line i, print a single integer that is the number of other cows patted by cow i.
5
2
1
2
3
4
2
0
2
1
3
The 5 cows are given the numbers 2, 1, 2, 3, and 4, respectively. The first cow pats the second and third cows; the second cows pats no cows; etc.
#include<stdio.h> int num1[1000100]={0}; int num2[1000100]={0}; int a[100100]; int main() { int n,i,mix=0,j; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&a[i]); num1[a[i]]++; } for(i=1;i<=n;i++) if(mix<a[i]) mix=a[i]; for(i=1;i<=mix;i++) { if(num1[i]!=0) for(j=i;j<=mix;j=j+i) { num2[j]=num2[j]+num1[i]; } } for(i=1;i<=n;i++) printf("%d\n",num2[a[i]]-1); }
标签：gif 技术 spl head 次数 lines hint div inpu
原文地址：https://www.cnblogs.com/shuaihui520/p/8975764.html
