Educational Codeforces Round 26 B,C

B. Flag of Berland

链接：http://codeforces.com/contest/837/problem/B

思路：题目要求判断三个字母是否是条纹型的，而且宽和高相同，那么先求出三个字母的边界，算下面积，是否和数量相同，不相同的话肯定不为条纹型，然后判断下他们宽和高的关系就行了

实现代码：

#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<list>
using namespace std;
#define ll long long
#define sd(x) scanf("%d",&x)
#define sdd(x,y) scanf("%d%d",&x,&y)
#define sddd(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define sf(x) scanf("%s",x)
#define ff(i,x,y) for(int i = x;i <= y;i ++)
#define fj(i,x,y) for(int i = x;i >= y;i --)
#define mem(s,x) memset(s,x,sizeof(s));
#define pr(x) printf("%d",x);
const int Mod = 1e9+;
const int inf = 1e9;
const int Max = 1e5+;
//void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=1;y=0;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}
//ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}
//int gcd(int a,int b)  {  return (b>0)?gcd(b,a%b):a;  }
//int lcm(int a, int b)  {  return a*b/gcd(a, b);   }    

int main()
{
    int n,m,en_rx=,be_rx=inf,en_ry=,be_ry=inf,en_gx=,en_gy=,en_bx=,en_by=,be_gx=inf,be_gy=inf,be_bx=inf,be_by=inf,r,g,b;
    char mp[][];
    sdd(n,m);
    r=;g=;b=;
    ff(i,,n){
        ff(j,,m){
            cin>>mp[i][j];
           if(mp[i][j]=='R'){
                r++; en_rx = max(i,en_rx);en_ry = max(j,en_ry);be_rx=min(i,be_rx);be_ry=min(j,be_ry);}
           if(mp[i][j]=='G'){ g++; en_gx = max(i,en_gx);en_gy = max(j,en_gy);be_gx=min(i,be_gx);be_gy=min(j,be_gy);}
           if(mp[i][j]=='B'){ b++; en_bx = max(i,en_bx);en_by = max(j,en_by);be_bx=min(i,be_bx);be_by=min(j,be_by);}
        }
    }
    if(n*m<){
        cout<<"NO"<<endl;return ;}
    int num = (en_rx - be_rx+)*(en_ry - be_ry+);
    //cout<<num<<endl;
    //cout<<en_gx<<" "<<be_gx<<endl;
    //cout<<en_ry<<" "<<be_ry<<endl;
    if((en_rx - be_rx)==(en_gx - be_gx)&&(en_gx - be_gx)==(en_bx - be_bx)&&(en_ry - be_ry)==(en_gy - be_gy)&&(en_gy - be_gy)==(en_by - be_by)&&num==r)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
     return ;
}

C. Two Seals

题目链接：http://codeforces.com/contest/837/problem/C

思路：

给你几个方块，让你选两个方块放在a*b范围里，要求面积最大，方块可以转90度，直接暴力模拟就是了

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll x[],y[];
int a,b;
ll max(ll x,ll y)
{
    if(x>y) return x;
    return y;
}
bool check(int i,int j){
            ll a1=x[i]+x[j],b1=max(y[i],y[j]);
            ll a2=y[i]+x[j],b2=max(x[i],y[j]);
            ll a3=x[i]+y[j],b3=max(y[i],x[j]);
            ll a4=y[i]+y[j],b4=max(x[i],x[j]);
            if((a1<=a && b1<=b) || (a2<=a && b2<=b) || (a3<=a && b3<=b) || (a4<=a && b4<=b))
            return true;
            if((a1<=b && b1<=a) || (a2<=b && b2<=a) || (a3<=b && b3<=a) || (a4<=b && b4<=a))
            return true;
            return false;
}
int main()
{
    int n,c=,y1,x1,i,j;
    cin>>n>>a>>b;
    for(i=;i<n;i++)
        cin>>x[i]>>y[i];
    ll ans = ,maxx = ;
    for(i=;i<n-;i++){
        for(j=i+;j<n;j++){
             if(check(i,j)==){
             ans = x[i]*y[i]+x[j]*y[j];
            //cout<<i<<" "<<j<<" "<<ans<<endl;
            maxx = max(maxx,ans);
            }
        }
    }
    cout<<maxx<<endl;
    return ;
}

ps:好鸡儿菜啊，打了一年还是这么水。