Croc Champ 2013 - Round 2 (Div. 2 Edition)-C Weird Gam

这题实在耗我很长时间。主要浪费在区间边界的特判。自然界的对称性很好，我们往往可以考虑特殊情况，处理了一半，另一半相似的推出。这题，比较容易猜出一个优先级。即{1,1},{0,1},{1,0}{0,0} 不影响答案根据{1,1}出现的次数的奇偶性等价成0,1然后统计{0,1}{1,0}出现次数特判一下吧。。纸上画一下。。这题有一个特点就是很多东西都互相抵消了，这就出现了一个常见的情况，余2为偶结果+1。意会吧。

今天重写，发现以前代码相当挫比。。。

/* * Author:  raylee * Created Time:  2014/1/21 15:13:17 * File Name: a.cpp *///#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<string>#include<map>#include<set>#include<vector>#include<queue>#include<stack>#include<ctime>#include<climits>using namespace std;#define pb push_back#define vi vector<int>#define clr( x , y ) memset(x, y, sizeof(x))typedef long long ll;char str1[ 2 * 1001000 ], str2[ 2 * 1001000 ];int main(){    ios::sync_with_stdio(false);    int n; cin>>n;    cin>>str1 + 1; cin>>str2 + 1;    int s11 = 0, s10 = 0, s01 = 0;    for(int i = 1; i <= 2 * n; i++){        if(str1[ i ] == '1' && str2[ i ] == '1') s11++;         else if(str1[ i ] == '0' && str2[ i ] == '1') s01++;        else if(str1[ i ] == '1' && str2[ i ] == '0') s10++;    }    int score_fir = 0, score_sec = 0;    score_fir = s11 - s11 / 2; score_sec = s11 / 2;    if(s11 % 2 == 1){        if(s01){            s01--; score_sec++;        }else if(s10){            s10--;        }        else {            cout<<"First"<<endl;            return 0;        }    }    if(s10 > s01){        score_fir += s01; score_sec += s01;        s10 -= s01;        int res = s10;        score_fir += res - res/2;    }else if(s10 < s01){        score_fir += s10; score_sec += s10;        s01 -= s10;        int res = s01;        score_sec += res/2;    }    if(score_fir > score_sec)        cout<<"First"<<endl;    else if(score_fir < score_sec)        cout<<"Second"<<endl;    else cout<<"Draw"<<endl;    return 0;}