#include <iostream>
using namespace std;

const int NUMBERS_PER_TICKET = 7;
const int NUM_PRIZES = 7;

int main() {
    int n;
    cin >> n;
    int winningNumbers[NUMBERS_PER_TICKET];
    // 读取中奖号码
    for (int i = 0; i < NUMBERS_PER_TICKET; ++i) {
        cin >> winningNumbers[i];
    }

    int prizeCounts[NUM_PRIZES] = {0};  // 初始化各奖项中奖张数为 0

    // 处理每张彩票
    for (int i = 0; i < n; ++i) {
        int ticketNumbers[NUMBERS_PER_TICKET];
        // 读取当前彩票的号码
        for (int j = 0; j < NUMBERS_PER_TICKET; ++j) {
            cin >> ticketNumbers[j];
        }

        int matchCount = 0;  // 记录当前彩票与中奖号码相同的数量
        // 统计相同号码的数量
        for (int j = 0; j < NUMBERS_PER_TICKET; ++j) {
            for (int k = 0; k < NUMBERS_PER_TICKET; ++k) {
                if (ticketNumbers[j] == winningNumbers[k]) {
                    ++matchCount;
                    break;
                }
            }
        }

        // 根据相同号码的数量确定中奖等级并更新对应奖项的中奖张数
        if (matchCount > 0) {
            prizeCounts[NUM_PRIZES - matchCount]++;
        }
    }

    // 输出各奖项的中奖张数
    for (int i = 0; i < NUM_PRIZES; ++i) {
        cout << prizeCounts[i]<<" ";
    }
    cout << endl;
    return 0;
}
这个题解是我的思路加上AI给的修改，下面两位学长的方法很好，是对我们输的号码，一个一个进行判断，但我的思路是把所有的号码全部输入完，然后进行两层for循环找到我们的数，然后又通过一个数组来储存答案，发现了中奖个数与数组下标的关系，然后进行储存，最后进行输出。

学长的方法

#include <stdio.h>

int main () {

int n;

scanf("%d", &n);

int a[7];

int b;

for (int i = 0; i < 7; i++) {

scanf("%d", &a[i]);

}

int t1 = 0, t2 = 0, t3 = 0, t4 = 0, t5 = 0, t6 = 0, t7 = 0;

for (int k = 0; k < n; k++) {

int t0 = 0;

for (int p = 0; p < 7; p++) {

scanf("%d", &b);

for (int q = 0; q < 7; q++) {

if (b == a[q]) {

t0++;

break;

}

}

}

if (t0 == 7) {

t1++;

} else if (t0 == 6) {

t2++;

} else if (t0 == 5) {

t3++;

} else if (t0 == 4) {

t4++;

} else if (t0 == 3) {

t5++;

} else if (t0 == 2) {

t6++;

} else if (t0 == 1) {

t7++;

}

}

printf("%d %d %d %d %d %d %d\n", t1, t2, t3, t4, t5, t6, t7);

return 0;

}

#include <stdio.h>
int main()
{
    int times,time;
    scanf("%d",&times);
    int a[9],b,i,j,t,r[9]={0};
    for(i=1;i<=7;i++)scanf("%d",&a[i]);
    for(time=1;time<=times;time++)
    {
        t=0;
        for(i=1;i<=7;i++)
        {
            scanf("%d",&b);
            for(j=1;j<=7;j++)
                if(b==a[j]){t++;break;}
        }
        r[t]++;
    }
    for(i=7;i>=1;i--)printf("%d ",r[i]);
    return 0;
}//