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http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1803
Solution:
考虑两个数x,y乘积%2016=0
x×y≡0(MOD 2016)
x=p1×2016+q1
y=p2×2016+q2
x×y=(p1×2016+q1)×(p2×2016+q2)=2016^2×p1p2+2016(p1q2+q1p2)+p1p2≡0(MOD 2016)
实际上就转化为余数乘积取模=0,预处理没两个余数乘积是否mod2016=0
统计答案两个余数出现的个数相乘即可(注意特判0不能选)
复杂度:O(2016^2)
// <1803.cpp> - Wed Oct 19 08:25:53 2016 // This file is made by YJinpeng，created by XuYike‘s black technology automatically. // Copyright (C) 2016 ChangJun High School, Inc. // I don‘t know what this program is. #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> #define MOD 2016 #define src(x,n) (n/MOD+(x!=0?(n%MOD>=x):0)) using namespace std; typedef long long LL; vector<pair<int,int> >a; int main() { freopen("1803.in","r",stdin); freopen("1803.out","w",stdout); for(int i=0;i<MOD;i++) for(int j=0;j<MOD;j++) if((i*j)%MOD==0) a.push_back(make_pair(i,j)); int n,m,to=a.size(); while(~scanf("%d%d",&n,&m)){ LL ans=0; for(int i=0;i<to;i++) ans+=1LL*src(a[i].first,n)*src(a[i].second,m); printf("%lld\n",ans); } return 0; }
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原文地址：http://www.cnblogs.com/YJinpeng/p/5978947.html
