1 solutions

  • 0
    @ 2024-12-5 18:22:03

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    #define N 10005
    #define M 105
    int n, m, ans;
    int x[N][M];//x[i][j]:第i层第j房间的标牌号,表示逆时针方向选择第x[i][j]房间可以上楼 
    int upNum[N];//upNum[i]:第i层有几个有楼梯的房间 
    int isUp[N][M];//isUp[i][j]:第i层第j房间是否有楼梯 
    int main()
    {
        scanf("%d %d", &n ,&m);
        for(int i = 1; i <= n; ++i)
            for(int j = 0; j < m; ++j)
            {
                scanf("%d %d", &isUp[i][j], &x[i][j]);
                if(isUp[i][j])//统计第i层有楼梯的房间数量
                    upNum[i]++;
            }
        int j, roomNum;//roomNum:需要选择有楼梯的房间的最少次数 
        scanf("%d", &j);//起始房间号 
        for(int i = 1; i <= n; ++i)
        {
            ans = (ans + x[i][j]) % 20123;//一边加一边取模
            if(x[i][j] % upNum[i] == 0)//求出需要选择房间的最少次数
                roomNum = upNum[i];
            else
                roomNum = x[i][j] % upNum[i];
            int k = 0;//看过的有楼梯的房间数
            if(isUp[i][j])
                k++;
            while(k < roomNum)//循环结束后,j指向第roomNum个房间,进入上一层的j房间 
            {
                j = (j + 1) % M;//循环遍历数组,M-1的下一个位置是0   
                if(isUp[i][j])//如果j房间有楼梯 
                    k++;
            }
        }
        cout << ans; 
    	return 0;
    }
    
    
    • 1

    Information

    ID
    9181
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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