题目216 A problem is easy
2016-07-30 08:42
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已AC代码:
#include<cstdio> main() { int n,i,t; scanf("%d",&t); while(t--) { int s=0; scanf("%d",&n); for(i=1;(i+1)*(i+1)<=n+1;i++) { if((n+1)%(i+1)==0) s++; } printf("%d\n",s); } }
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