【codeforces】Divisibility
2016-07-23 11:49
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只要知道n/k为n一下能被k整除的数的总和就行了。
#include<stdio.h> typedef long long ll; ll abs(ll x) { return x>0?x:-x; } int main() { ll k,a,b; while(~scanf("%I64d%I64d%I64d",&k,&a,&b)) { ll ans; if(a>0&&b>0) { ans=abs(b/k)-abs((a-1)/k); } else if(a<0&&b<0) { ans=abs(a/k)-abs((b+1)/k); } else { ans=abs(a/k)+abs(b/k)+1; } printf("%I64d\n",ans); } return 0; }
http://acm.hust.edu.cn/vjudge/contest/123213#problem/A
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