CSU 1803 2016 （逆向思维）

让你在[1,n]和[1,m]中选一个a和一个b，使得a×b是2016的倍数
问你有多少种情况，然后直接考虑可能比较麻烦
这题就要反过来考虑，就是考虑不是2016的倍数有多少种
a×b%2016!=0
a%2016!=0并且b%2016!=0
那么直接枚举a和b模2016的余数就行了
a%2016=i,b%2016=j
i×j%2016!=0
ans=∑∑(n−i2016+1)×(m−j2016+1)

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代码：

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <string>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <sstream>#include <cstdlib>#include <iostream>#include <algorithm>#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;#define   MAX           1000005#define   MAXN          1000005#define   maxnode       205#define   sigma_size    2#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   lrt           rt<<1#define   rrt           rt<<1|1#define   middle        int m=(r+l)>>1#define   LL            long long#define   ull           unsigned long long#define   mem(x,v)      memset(x,v,sizeof(x))#define   lowbit(x)     (x&-x)#define   pii           pair<int,int>#define   bits(a)       __builtin_popcount(a)#define   mk            make_pair#define   limit         10000//const int    prime = 999983;const int    INF   = 0x3f3f3f3f;const LL     INFF  = 0x3f3f;//const double pi    = acos(-1.0);const double inf   = 1e18;const double eps   = 1e-9;const LL     mod   = 1e9+7;const ull    mx    = 133333331;/*****************************************************/inline void RI(int &x) {      char c;      while((c=getchar())<'0' || c>'9');      x=c-'0';      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0'; }/*****************************************************/int main(){    //freopen("in.txt","r",stdin);    int n,m;    while(cin>>n>>m){        LL ans=(LL)n*m;        for(int i=1;i<=min(2015,n);i++){            for(int j=1;j<=min(2015,m);j++){                if(i*j%2016!=0){                    ans-=(LL)((n-i)/2016+1)*((LL)(m-j)/2016+1);                }            }        }        cout<<ans<<endl;    }    return 0;}