Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E: Prerequisites?

Freddie the frosh has chosen to take k courses. To meet the degree requirements, he must take courses from each of several categories. Can you assure Freddie that he will graduate, based on his course selection?

Input consists of several test cases. For each case, the first line of input contains 1 ≤ k ≤ 100, the number of courses Freddie has chosen, and 0 ≤ m ≤ 100, the number of categories. One or more lines follow containing k 4-digit integers follow; each is the number of a course selected by Freddie. Each category is represented by a line containing 1 ≤ c ≤ 100, the number of courses in the category, 0 ≤ r ≤ c, the minimum number of courses from the category that must be taken, and the c course numbers in the category. Each course number is a 4-digit integer. The same course may fulfil several category requirements. Freddie's selections, and the course numbers in any particular category, are distinct. A line containing 0 follows the last test case.

For each test case, output a line containing "yes" if Freddie's course selection meets the degree requirements; otherwise output "no."

Sample Input

3 2
0123 9876 2222
2 1 8888 2222
3 2 9876 2222 7654 
3 2
0123 9876 2222
2 2 8888 2222
3 2 7654 9876 2222
0

Output for Sample Input

yes
no

---

Gordon V. Cormack


詢問能不能符合畢業門檻, 畢業門檻就是分類中至少要選幾門課。
然後已經知道他已經過了多少門課的課號。

#include <stdio.h>
int main() {
    int n, m, i, x, r, c;
    while(scanf("%d %d", &n, &m) == 2) {
        int mark[10000] = {};
        for(i = 0; i < n; i++)
            scanf("%d", &x), mark[x] = 1;
        int ok = 1;
        for(i = 0; i < m; i++) {
            scanf("%d %d", &r, &c);
            int cnt = 0;
            while(r--) {
                scanf("%d", &x);
                cnt += mark[x];
            }
            ok &= cnt >= c;
        }
        puts(ok ? "yes" : "no");
    }
    return 0;
}