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  • poj 2469 Stack 'em Up

    最近做挑战编程,题目难度加大,题意理解也越来越吃力了,好几次都理解错题意。

    题意:先给出洗牌者能够洗牌的几种方法,再给你k,让你求他经过第k种方法洗牌后的排序

    理解:每次下面给出的k使每一次都重新洗一次,而不是让你输出第k组数据

    #include<stdio.h>
    int main()
    {
        int n,k,i,a,j;
        char str[15][10]= {"Ace","2","3","4","5","6","7","8","9","10","Jack","Queen","King"};
        int t[110][110];//储存的命令
        int save[110][60];
        scanf("%d",&n);
        for (i=1; i<=52; i++)
            save[0][i]=i;
        for (i=1; i<=n; i++)
            for (j=1; j<=52; j++)
                scanf("%d",&t[i][j]);
        while(scanf("%d",&a)!=EOF)
        {
            for (i=1; i<=52; i++)
            {
                //i是位置j,t[i]整数i
                save[a][i]=save[0][t[a][i]];
            }
            for (i=1;i<=52;i++)
            {
                save[0][i]=save[a][i];
            }
            for (i=1; i<=52; i++)
            {
                //printf("%d ",save[a][i]);
                if(save[a][i]>=1 && save[a][i]<=13) printf("%s of Clubs\n",str[save[a][i]%13]);
                else if(save[a][i]>=14 && save[a][i]<=26) printf("%s of Diamonds\n",str[save[a][i]%13]);
                else if(save[a][i]>=27 && save[a][i]<=39) printf("%s of Hearts\n",str[save[a][i]%13]);
                else if(save[a][i]>=40 && save[a][i]<=52) printf("%s of Spades\n",str[save[a][i]%13]);
            }
            printf("\n");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zsboy/p/2511084.html
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