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 Description
Company A has several production lines. These production lines have a
global timer for synchronization. Workers feed in raw materials into a mixer
machine, and then the mixed material is dispatched to these production lines
so as to yield products. Every product has a unique product number for
quality control. A product number has three elds of the form ppp-ttttttt-q,
where ppp is a three-digit number representing the production line, ttttttt is
a seven-digit number representing the manufacture time (in seconds), and q
is a single digit for indicating the quality of this product.
The product may have defects due to a malfunction of a machine or
an incomplete mixture of the raw material. If a machine malfunctions, the
defective rate of the production line containing this malfunctioning machine
becomes high. On the other hand, once the raw material is not mixed well,
the defective rates of all production lines rise. Write a program that reads a
series of product numbers and outputs an alarm message instantly if one or
both of the following two types of event occur.
 The rst type of event is that the number of defective products yielded
by a single production line in t1 seconds is greater than or equal to k1.
 The second type of event is that the total number of defective products
yielded by all production lines in t2 seconds is greater than or equal to
k2.

---

看懂題目其實就很簡單了, 是在 t 秒內, 並不是 0~t 秒,
英文比較容易卡這個, 剩下的就不說了


#include <stdio.h>
#include <stdlib.h>
typedef struct {
    int p, t, q;
}DD;
DD D[10000];
int cmp(const void *i, const void *j) {
    DD *a, *b;
    a = (DD *)i, b = (DD *)j;
    return a->t - b->t;
}
int main() {
    int t, t1, t2, k1, k2, n;
    int i, p, tt, q;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d %d %d %d", &t1, &t2, &k1, &k2, &n);
        for(i = 0; i < n; i++) {
            scanf("%d-%d-%d", &p, &tt, &q);
            D[i].p = p, D[i].t = tt, D[i].q = q;
        }
        qsort(D, n, sizeof(DD), cmp);
        int mm[1000] = {}, flag = 0, j = 0;
        for(i = 0; i < n; i++) {
            while(j < n && D[j].t < D[i].t-t1) {
                if(D[j].q == 1)
                    mm[D[j].p]--;
                j++;
            }
            if(D[i].q == 1) {
                mm[D[i].p]++;
                if(mm[D[i].p] >= k1)
                    flag = 1;
            }
        }
        j = 0;
        int total = 0;
        for(i = 0; i < n; i++) {
            while(j < n && D[j].t < D[i].t-t2) {
                if(D[j].q == 1)
                    total--;
                j++;
            }
            if(D[i].q == 1) {
                total++;
                if(total >= k2)
                    flag = 1;
            }
        }
        puts(flag ? "ALARM" : "GOOD");
    }
    return 0;
}
/*
3
2 1 2 3 6
000-0000000-1
000-0000001-1
001-0000000-0
001-0000002-0
003-0000003-0
001-0000000-0
2 1 2 3 6
000-0000000-0
000-0000001-1
001-0000000-0
001-0000001-1
003-0000001-1
006-0000001-0
2 1 2 3 6
000-0000000-0
000-0000001-1
001-0000000-0
005-0000001-0
001-0000009-0
011-0000001-0
*/