题目链接:https://vjudge.net/contest/350917#problem/A
思路:
分情况很复杂,考虑一下暴力(当时怎么就没想到呢?)
暴力枚举第一个分区中a = 1,b = 0的情况设为i,和a = 1,b = 1的情况设为j。
然后计算出相关的变量判断是否合法,要先求出b2,然后再求出其他变量。
c | a | first | second |
0 | 0 | x | y |
1 | 0 | i | a[1][0] - i |
0 | 1 | a[0][1] - b2 | b2 = i+j - (a[1][0] - i) - (a[1][1] - j) |
1 | 1 | j | a[1][1] - j |
只要判断相关的值是否合法就好了,
最后算出x = n/2 - (i + j + a[0][1] - b2)
然后输出位置就好了。
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
int a[5][5] = {0},n;
char s1[N],s2[N];
int main(void)
{
scanf("%d",&n);
scanf("%s%s",s1+1,s2+1);
for(int i=1;i<=n;i++){
a[ s1[i] - '0' ][ s2[i] - '0' ]++;
}
int t1 = -1,t2 = -1,t3 = -1,t0 = -1;
bool fg = false;
for(int i=0;i<=a[1][0];i++)
{
for(int j=0;j<=a[1][1];j++)
{
int b2 = i+j-a[1][1]+j;
if(i+j>n/2 || b2<0 || b2>a[0][1]) continue;
int aa = i+j+a[0][1] - b2;
int bb = (a[1][0]-i) + b2 + (a[1][1]-j);
if(aa>n/2 || bb>n/2) continue;
t0 = n/2-aa;
t1 = i;
t2 = a[0][1] - b2;
t3 = j;
fg = true;break;
}
if(fg == true) break;
}
if(fg == false)
{
printf("-1\n");
return 0;
}
//printf("t1 = %d,t2 = %d,t3 = %d,t0 = %d\n",t1,t2,t3,t0);
for(int i=1;i<=n;i++)
{
if(t0>0 && s1[i] == '0' && s2[i] == '0') printf("%d ",i),t0--;
if(t1>0 && s1[i] == '1' && s2[i] == '0') printf("%d ",i),t1--;
if(t2>0 && s1[i] == '0' && s2[i] == '1') printf("%d ",i),t2--;
if(t3>0 && s1[i] == '1' && s2[i] == '1') printf("%d ",i),t3--;
}
return 0;
}