彩票统计(oj题)

2023-12-19 22:59:46

最近要期末考了,真的很忙

发这个文主要是来炫耀的

唯一一个全对的

代码如下:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int cmp(const void * a, const void * b)
{
    return * ((int *) a) - * ((int *) b);
}
struct Ticket{
    int l[6];
    int n;
}ticket[1000000];
int t[10][10][10][10][10][10], count;
struct Ticket * p[1000000];

int main(void)
{
    int N;
    scanf("%d", &N);
    for(int i = 0; i < N; i++)
    {
        int tmp[6];
        for(int j = 0; j < 6; j++)
            scanf("%d", &tmp[j]);
        qsort(tmp, 6, sizeof(int), cmp);
        t[tmp[0]][tmp[1]][tmp[2]][tmp[3]][tmp[4]][tmp[5]]++;
    }
    for(int x1 = 0; x1 < 10; x1++)
        for(int x2 = x1; x2 < 10; x2++)
            for(int x3 = x2; x3 < 10; x3++)
                for(int x4 = x3; x4 < 10; x4++)
                    for(int x5 = x4; x5 < 10; x5++)
                        for(int x6 = x5; x6 < 10; x6++)
                        if(t[x1][x2][x3][x4][x5][x6])
                        {
                            ticket[count].l[0] = x1, ticket[count].l[1] = x2, ticket[count].l[2] = x3;
                            ticket[count].l[3] = x4, ticket[count].l[4] = x5, ticket[count].l[5] = x6;
                            ticket[count++].n = t[x1][x2][x3][x4][x5][x6];
                        }
    for(int i = 0; i < count; i++)
        p[i] = &ticket[i];
    for(int i = 0; i < count - 1; i++)
        for(int j = i + 1; j < count; j++)
        {
            if(p[i]->n < p[j]->n)
            {
                struct Ticket * tmp = p[i];
                p[i] = p[j];
                p[j] = tmp;
            }
            else if(p[i]->n == p[j]->n)
            {
                for(int k = 0; k < 6; k++)
                {
                    if((p[i]->l)[k] > (p[j]->l)[k])
                    {
                        struct Ticket * tmp = p[i];
                        p[i] = p[j];
                        p[j] = tmp;
                        break;
                    }
					else if((p[i]->l)[k] < (p[j]->l)[k])
						break;
                }
            }
        }
    for(int i = 0; i < count; i++)
    {
        for(int j = 0; j < 6; j++)
            printf("%d ", (p[i]->l)[j]);
        printf("%d\n", p[i]->n);
    }
    return 0;
}

文章来源:https://blog.csdn.net/Fool256353/article/details/135095060
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