題解

題目大意 n個操作 基礎值r 定義五種操作 add將r增加v beq看r是否等於v如果等於則跳轉到第k條指令 bne看r是否不等於v如果不等於則跳轉到第k條指令 blt看r是否嚴格小於v如果小於則跳轉到第k條指令 bgt看r是否嚴格大於v如果大於則跳轉到第k條指令 指令編號從1到n

模擬一下 如果執行同一條指令前r的值出現過則出現死迴圈輸出No 否則迴圈超出n輸出Yes

AC程式碼

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long
 
 long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 1e4 + 10;
int o[MAXN], v[MAXN], k[MAXN];
bool vis[MAXN][260];
char s[100];

int main()
{
#ifdef LOCAL
    freopen("C:/input.txt", "r", stdin);
#endif
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        scanf("%d", &n);
        for
 
 (int i = 1; i <= n; i++)
        {
            scanf("%s%d", s, &v[i]);
            if (strcmp(s, "add") == 0)
                o[i] = 1;
            else
            {
                scanf("%d", &k[i]);
                if (strcmp(s, "beq") == 0)
                    o[i] = 2;
                else
 
 if (strcmp(s, "bne") == 0)
                    o[i] = 3;
                else if (strcmp(s, "blt") == 0)
                    o[i] = 4;
                else if (strcmp(s, "bgt") == 0)
                    o[i] = 5;
            }
            memset(vis[i], 0, sizeof(vis[i]));
        }
        int p = 1, r = 0;
        while (p <= n)
        {
            if (vis[p][r])
                break;
            vis[p][r] = true;
            if (o[p] == 1)
                r = (r + v[p]) % 256, p++;
            else if (o[p] == 2)
            {
                if (r == v[p])
                    p = k[p];
                else
                    p++;
            }
            else if (o[p] == 3)
            {
                if (r != v[p])
                    p = k[p];
                else
                    p++;
            }
            else if (o[p] == 4)
            {
                if (r < v[p])
                    p = k[p];
                else
                    p++;
            }
            else if (o[p] == 5)
            {
                if (r > v[p])
                    p = k[p];
                else
                    p++;
            }
        }
        if (p > n)
            printf("Yes\n");
        else
            printf("No\n");
    }

    return 0;
}