UOJ454. 【UER 8】打雪仗

UOJ454. 【UER 8】打雪仗

对于每三个数，bob 告诉 alice 前两个数是否需要，alice 告诉 bob 需要的数并直接告诉第 3 个数，这样 bob 的次数为 $2n \times \frac 2 3 = \frac 4 3 n$ ，alice 的次数为 $n + n \times \frac 1 3 = \frac 4 3 n$ ，这样都满足条件了。

千万不要用 scanf("%1d", &a) ，被卡了一个小时不知道哪里错了，大概是 scanf 会一直等，然后 T 飞。

查看代码

//alice.cpp
#include <cstdio>
#include <cstdlib>
#include <iostream>
using namespace std;
const int N = 1e3 + 10;
int n, m, s[N << 1];
int main ()
{
    FILE *in = fopen("alice.in", "r");
    fscanf(in, "%d%d", &n ,&m);
    for (int i = 1; i <= n << 1; ++i) fscanf(in, "%1d", &s[i]);
    for (int i = 1; i <= n << 1; i += 3)
    {
        if (getchar() == '1') printf("%d", s[i]);
        if (getchar() == '1') printf("%d", s[i + 1]);
        printf("%d", s[i + 2]);
        fflush(stdout);
    }
    return 0;
}
//bob.cpp
#include <cstdio>
#include <vector>
using namespace std;
const int N = 1e3 + 10;
vector <bool> ans;
int n, m;
bool p[N << 1];
int main ()
{
    FILE *in = fopen("bob.in", "r"), *out = fopen("bob.out", "w");
    fscanf(in, "%d%d", &n, &m);
    for (int i = 1, t; i <= n; ++i)
        fscanf(in, "%d", &t), p[t] = true;
    for (int i = 1; i <= n << 1; i += 3)
    {
        printf("%d%d", p[i], p[i + 1]);
        fflush(stdout);
        if (p[i]) ans.push_back(getchar() - '0');
        if (p[i + 1]) ans.push_back(getchar() - '0');
        bool a = getchar() - '0'; if (p[i + 2]) ans.push_back(a);
    }
    for (bool i : ans) fprintf(out, "%d", i);
    return 0;
}