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
Koerner's Pub
Time Limit: 1 second

"Homer no function beer well without."

Homer Simpson

Every Friday night, Alice, Clara and Mary go to Koerner's pub to relax after a long week. At the pub, lots of guys ask them for their phone numbers. In fact, the three ladies are so popular that they have started counting the number of times each one is asked for her phone number during one evening. On day i, Alice, Clara and Mary were asked A_i, C_i and M_i times, respectively. 100 Fridays have passed, and the records were lost, but there are 3 things Alice still remembers.

1. X of the 100 days, A_i was equal to C_i.
2. Y of the 100 days, C_i was equal to M_i.
3. Z of the 100 days, A_i was equal to M_i.

How many time were A_i, C_i and M_i all equal?

Input
The first line of input gives the number of cases, N. N test cases follow. Each one contains the integers X, Y and Z on a line.

Output
For each test case, output one line containing "Case #x:" followed by "Between A and B times.", where A and B are the lowest and highest number of times that A_i, C_i and M_i could have been equal. If the situation is impossible, print "The records are faulty." instead.

Sample Input	Sample Output
5 50 50 50 100 100 99 100 50 50 101 50 50 100 100 100	Case #1: Between 25 and 50 times. Case #2: The records are faulty. Case #3: Between 50 and 50 times. Case #4: The records are faulty. Case #5: Between 100 and 100 times.

---

Problemsetter: Igor Naverniouk
Alternate solution: Yury Kholondyrev


題目描述：

三個人，在 100 天內，在每一天都有不同個數人向這三個人要電話號碼，
告訴你，他們有兩兩具有 X, Y, Z 天相等，問有多少天三個人都具有相同的數量。

題目解法：

這是一個不等式的計算。

考慮最少天數
假設相同天數 w, 對於 Mary 來說, 只與 Alice 有 z-w 天, 只與 Clara 有 y-w 天，
對於 Clara 來說, 存在 x-w 天是單獨與 Alice, 因此在 Mary 看來是不會有任何一天相同。

則對於 Mary 來說, (x-w)+(y-w)+(z-w)+w <= 100, 便可得到 w。

考慮最多天即 min(x, y, z)

#include <stdio.h>
#include <algorithm>
using namespace std;

int main() {
    int testcase, cases = 0;
    int x, y, z;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d", &x, &y, &z);
        int r = min(x, min(y, z));
        int l = (x+y+z-100+1)/2;//ceil
        if(x+y+z <= 100)
            l = 0;
        if(l <= r && l >= 0)
            printf("Case #%d: Between %d and %d times.\n", ++cases, l, r);
        else
            printf("Case #%d: The records are faulty.\n", ++cases);
    }
    return 0;
}