poj1426之bfs打表解法

题意：给你个数n，求出一个n的倍数，满足只由1和0组成

分析：由1,10,11,101,111...往后bfs，发现Memory Limit Exceed。然后打表，轻松ac。

打表：

#include <queue>
#include <cstdio>
using namespace std;

typedef long long ll;
int n;
queue<ll> qu;

void bfs()
{
while (!qu.empty()){
ll ans = qu.front();qu.pop();
if(ans % n == 0){
printf("%lld,\n",ans);
return;
}
qu.push(ans*10);
qu.push(ans*10+1);
}
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif

for (int i = 1; i < 201; i += 1){
n = i;
while (!qu.empty()) qu.pop();
qu.push(1);
bfs();
}
return 0;
}

然后打表：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

typedef long long ll;
int n;
ll ans[200] = {
1,
10,
111,
100,
10,
1110,
1001,
1000,
111111111,
10,
11,
11100,
1001,
10010,
1110,
10000,
11101,
1111111110,
11001,
100,
10101,
110,
110101,
111000,
100,
10010,
1101111111,
100100,
1101101,
1110,
111011,
100000,
111111,
111010,
10010,
11111111100,
111,
110010,
10101,
1000,
11111,
101010,
1101101,
1100,
1111111110,
1101010,
10011,
1110000,
1100001,
100,
100011,
100100,
100011,
11011111110,
110,
1001000,
11001,
11011010,
11011111,
11100,
100101,
1110110,
1111011111,
1000000,
10010,
1111110,
1101011,
1110100,
10000101,
10010,
10011,
111111111000,
10001,
1110,
11100,
1100100,
1001,
101010,
10010011,
10000,
1111111101,
111110,
101011,
1010100,
111010,
11011010,
11010111,
11000,
11010101,
1111111110,
1001,
11010100,
10000011,
100110,
110010,
11100000,
11100001,
11000010,
111111111111111111,
100,
101,
1000110,
11100001,
1001000,
101010,
1000110,
100010011,
110111111100,
1001010111,
110,
111,
10010000,
1011011,
110010,
1101010,
110110100,
10101111111,
110111110,
100111011,
111000,
11011,
1001010,
10001100111,
11101100,
1000,
11110111110,
11010011,
10000000,
100100001,
10010,
101001,
11111100,
11101111,
11010110,
11011111110,
11101000,
10001,
100001010,
110110101,
100100,
10011,
100110,
1001,
1111111110000,
11011010,
100010,
1100001,
11100,
110111,
11100,
1110001,
11001000,
10111110111,
10010,
1110110,
1010100,
10101101011,
100100110,
100011,
100000,
11101111,
11111111010,
1010111,
1111100,
1111110,
1010110,
11111011,
10101000,
10111101,
111010,
1111011111,
110110100,
1011001101,
110101110,
100100,
110000,
100101111,
110101010,
11010111,
11111111100,
1001111,
10010,
100101,
110101000,
1110,
100000110,
1001011,
1001100,
1010111010111,
110010,
11101111,
111000000,
11001,
111000010,
101010,
110000100,
1101000101,
1111111111111111110,
111000011,
1000
};

int main()
{
while (cin >> n,n)
cout << ans[n-1] << endl;
return 0;
}