Codeforces Round #258 (Div. 2)C(暴力枚舉)

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就枚舉四種情況，哪種能行就是yes了。很簡單，關鍵是寫法，我寫的又醜又長。。。看了zhanyl的寫法頓時心生敬佩。寫的乾淨利落，簡直美如畫。。。這是功力的體現！

以下是zhanyl的寫法，轉載在此以供學習：

#include <vector>#include <list>#include <queue>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <tr1/unordered_set>#include <tr1/unordered_map>using namespace std;using namespace tr1;const int g[4][2]={{1,1},{1,-1},{-1,1},{-1,-1}};long long n,k,d[2],f[3];int t;bool ans;int main(){    scanf("%d",&t);    while(t--){        scanf("%I64d%I64d%I64d%I64d",&n,&k,&d[0],&d[1]);        if(n%3){            puts("no");            continue;        }        n=n/3;        ans=false;        for(int i=0;i<4;i++)            if((k-g[i][0]*d[0]-g[i][1]*d[1])%3==0){                f[1]=(k-g[i][0]*d[0]-g[i][1]*d[1])/3;                f[0]=f[1]+g[i][0]*d[0];                f[2]=f[1]+g[i][1]*d[1];                if(f[0]>=0&&f[1]>=0&&f[2]>=0&&f[0]<=n&&f[1]<=n&&f[2]<=n)ans=true;            }        if(ans)puts("yes");        else puts("no");    }    return 0;}

 

Codeforces Round #258 (Div. 2)C(暴力枚舉)