Codeforces Round #258 (Div. 2) C. Predict Outcome of the Game

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首先，如果n%3 != 0，肯定不会使得三个队伍平手，输出“no”。

设三个队在前K场比赛中分别赢了x1,x2,x3场，有|x1 - x2| = d1, | x2 - x3| = d2。

由于d1,d2分别有正负，那么共分为四种情况：

{x1 - x2 = d1, x2 - x3 = d2}，{x1 - x2 = d1, x3 - x2 = d2}，{x2 - x1 = d1, x2 - x3 = d2}，{x2 - x1 = d1,x3 - x2 = d2}。

我们分别解出这四个方程中的符合题意的解，若要使最后三个队赢的次数相同，只要x1 <= n/3 && x2 <= n/3 && x3 <= n/3即可，前提是它们都大于等于0。

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}int main()  {      int T;      ll n,k,d1,d2;      ll x1,x2,x3;        read(T);      while(T--)      {          read(n);read(k);read(d1)&read(d2);          if(n%3!=0)          {              printf("no\n");              continue;          }          int flag=0;          for(int i=0;i<=3;i++)          {              if(i==0)              {                  if((k+d2+2*d1)>=0&&(k+d2+2*d1)%3==0)                  {                      x1=(k+d2+2*d1)/3;                      x2=x1-d1;                      x3=x2-d2;                  }                  else continue;              }              else if(i==1)              {                  if((k-d2+2*d1)>=0&&(k-d2+2*d1)%3==0)                  {                      x1=(k-d2+2*d1)/3;                      x2=x1-d1;                      x3=x2+d2;                  }                  else continue;              }              else if(i==2)              {                  if((k+d2-2*d1)>=0&&(k+d2-2*d1)%3==0)                  {                      x1=(k+d2-2*d1)/3;                      x2=x1+d1;                      x3=x2-d2;                  }                  else continue;              }              else if(i==3)              {                  if((k-d2-2*d1)>=0&&(k-d2-2*d1)%3==0)                  {                      x1=(k-d2-2*d1)/3;                      x2=x1+d1;                      x3=x2+d2;                  }                  else continue;              }              if(x1>=0&&x2>=0&&x3>=0&&x1<=n/3&&x2<=n/3&&x3<= n/3)                  flag=1;          }          if(flag)              printf("yes\n");          else              printf("no\n");      }      return 0;  }

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