题目：
Yaroslav, Andrey and Roman can play cubes for hours and hours. But the game is for three, so when Roman doesn't show up, Yaroslav and Andrey play another game.
Roman leaves a word for each of them. Each word consists of 2·n binary characters "0" or "1". After that the players start moving in turns. Yaroslav moves first. During a move, a player must choose an integer from 1 to 2·n, which hasn't been chosen by anybody up to that moment. Then the player takes a piece of paper and writes out the corresponding character from his string.
Let's represent Yaroslav's word as s = s1s2... s2n. Similarly, let's represent Andrey's word as t = t1t2... t2n. Then, if Yaroslav choose number k during his move, then he is going to write out character sk on the piece of paper. Similarly, if Andrey choose number r during his move, then he is going to write out character tron the piece of paper.
The game finishes when no player can make a move. After the game is over, Yaroslav makes some integer from the characters written on his piece of paper (Yaroslav can arrange these characters as he wants). Andrey does the same. The resulting numbers can contain leading zeroes. The person with the largest number wins. If the numbers are equal, the game ends with a draw.
You are given two strings s and t. Determine the outcome of the game provided that Yaroslav and Andrey play optimally well.
Input
The first line contains integer n (1 ≤ n ≤ 106). The second line contains string s— Yaroslav's word. The third line contains string t — Andrey's word.
It is guaranteed that both words consist of 2·n characters "0" and "1".
Output
Print "First", if both players play optimally well and Yaroslav wins. If Andrey wins, print "Second" and if the game ends with a draw, print "Draw". Print the words without the quotes.
Examples
Input
2 0111 0001
Output
First
Input
3 110110 001001
Output
First
Input
3 111000 000111
Output
Draw
Input
4 01010110 00101101
Output
First
Input
4 01100000 10010011
Output
Second
解题报告： 给定两个字符串，就是分别代表1 0，每次双方都优先用1进行对决，且每次这个位两个人只能选择一个，那么就是贪心，每次都用自己的0 去堵死对方的1，以使自己的1 能更可能的留到最后。
ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e7+100;
char a[maxn],b[maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
n*=2;
cin>>a>>b;
int ans1=0,ans2=0,ans3=0;
for(int i=0;i<n;i++)
{
if(a[i]=='1'&&b[i]=='1')
ans1++;
if(a[i]=='1'&&b[i]=='0')
ans2++;
if(a[i]=='0'&&b[i]=='1')
ans3++;
}
ans1%=2;
if(ans1+ans2==ans3-1)
ans3--;
if(ans1+ans2==ans3)
printf("Draw\n");
else if(ans1+ans2>ans3)
printf("First\n");
else
printf("Second\n");
}
}