123减法

题目大意：给你三个数n，n1，n2，n3。你可以对n做下面三种操作之一：

• 将n减1

• 将n减2

• 将n减3

最多做100次操作，且n不能等于n1，n2，n3中的任意一个，问你是否可以将n变成0。

因为这题只有三种转移状态，我第一反应想到的是bfs搜索，很好实现，代码如下:

//https://blog.csdn.net/hesorchen
// #include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <cstring>
using namespace std;
#define ll long long
#define endl "\n"
#define mod 1000000007

int n, n1, n2, n3;
int t[308];
struct node
{
    int zhi, step;
} a, b;
queue<node> q;
int bfs()
{
    a.zhi = n;
    a.step = 0;
    q.push(a);
    while (!q.empty())
    {
        a = q.front();
        q.pop();
        if (a.step > 100 || a.zhi == n1 || a.zhi == n2 || a.zhi == n3||t[a.zhi])
            continue;
    t[a.zhi]=1;
        if (a.zhi == 0 && a.step <= 100)
            return 1;
        if (a.zhi - 1 >= 0)
        {

            b.zhi = a.zhi - 1;
            b.step = a.step + 1;
            q.push(b);
        }
        if (a.zhi - 2 >= 0)
        {
            b.zhi = a.zhi - 2;
            b.step = a.step + 1;
            q.push(b);
        }
        if (a.zhi - 3 >= 0)
        {
            b.zhi = a.zhi - 3;
            b.step = a.step + 1;
            q.push(b);
        }
    }
    return 0;
}
int main()
{
    cin >> n >> n1 >> n2 >> n3;
    if (bfs())
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}

赛后有人贴出代码之后发现这其实是贪心+模拟，因为有次数限制，所以每一次都尽量多减一点。

AC代码：

//https://blog.csdn.net/hesorchen
// #include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define ll long long
#define endl "\n"
#define mod 1000000007
int n, a, b, c;
int check(int x)
{
    if (x != a && x != b && x != c)
        return 1;
    return 0;
}
int main()
{
    cin >> n >> a >> b >> c;
    if (!a || !b || !c || !check(n))
        return !printf("NO\n");
    int flag = 1;
    int step = 0;
    while (n > 0)
    {
        step++;
        if (check(n - 3))
            n -= 3;
        else if (check(n - 2))
            n -= 2;
        else if (check(n - 1))
            n -= 1;
        else
        {
            flag = 0;
            break;
        }
    }
    if (flag && step <= 100)
        return !printf("YES\n");
    printf("NO\n");
    return 0;
}