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

Martin and Isa stopped playing crazy games and finally got married. It's good news! They're pursuing a new life of happiness for both and, moreover, they're moving to a new house in a remote place, bought with most of their savings.

Life is different in this new place. In particular, electricity is very expensive, and they want to keep everything under control. That's why Martin proposed to keep a daily record of how much electricity has been consumed in the house. They have an electricity meter, which displays a number with the amount of KWh (kilowatt-hour) that has been consumed since their arrival.

At the beginning of each day they consult the electricity meter, and write down the consumption. Some days Martin does it, and some days Isa does. That way, they will be able to look at the differences of consumption between consecutive days and know how much has been consumed.

But some days they simply forget to do it, so, after a long time, their register is now incomplete. They have a list of dates and consumptions, but not all of the dates are consecutive. They want to take into account only the days for which the consumption can be precisely determined, and they need help.

Input

The input contains several test cases. The first line of each test case contains one integer N indicating the number of measures that have been taken (2 ≤ N ≤ 10³). Each of the N following lines contains four integers D, M, Y and C, separated by single spaces, indicating respectively the day (1 ≤ D ≤ 31), month (1 ≤ M ≤ 12), year (1900 ≤ Y ≤ 2100), and consumption (0 ≤ C ≤ 10⁶) read at the beginning of that day. These N lines are increasingly ordered by date, and may include leap years. The sequence of consumptions is strictly increasing (this is, no two different readings have the same number). You may assume that D, M and Y represent a valid date.

Remember that a year is a leap year if it is divisible by 4 and not by 100, or well, if the year is divisible by 400.

The end of input is indicated by a line containing only one zero.

Output

For each test case in the input, your program must print a single line containing two integers separated by a single space: the number of days for which a consumption can be precisely determined, and the sum of the consumptions for those days.

5
9 9 1979 440
29 10 1979 458
30 10 1979 470
1 11 1979 480
2 11 1979 483
3
5 5 2000 6780
6 5 2001 7795
7 5 2002 8201
8
28 2 1978 112
1 3 1978 113
28 2 1980 220
1 3 1980 221
5 11 1980 500
14 11 2008 600
15 11 2008 790
16 12 2008 810
0

2 15
0 0
2 191

ACM ICPC :: South American Regional 2008

題目描述：

抄電表知道每日用電量，但是由於健忘，並不是每天都抄，
只有在兩天都有抄的情況下，才能知道前一天的用電量，問能確定的每日用電量總和。

題目解法：

模擬日期計算。

#include <stdio.h>
void addDay(int &mm, int &dd, int &yy) {
    int m[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int leap = (yy%4 == 0 && yy%100 != 0 || yy%400 == 0);
    if(leap)    m[2]++;
    dd++;
    if(dd > m[mm])  mm++, dd = 1;
    if(mm == 13)
        yy++, dd = 1, mm = 1;
}
int main() {
    int n, i;
    int mm[2000], yy[2000], dd[2000], cc[2000];
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++)
            scanf("%d %d %d %d", &dd[i], &mm[i], &yy[i], &cc[i]);
        int days = 0, cost = 0;
        for(i = 0; i < n-1; i++) {
            addDay(mm[i], dd[i], yy[i]);
            if(mm[i] == mm[i+1] && dd[i] == dd[i+1] && yy[i] == yy[i+1])
                days++, cost += cc[i+1]-cc[i];
        }
        printf("%d %d\n", days, cost);
    }
    return 0;
}

您可能對以下文章有興趣

[UVA] 498 - Polly the Polynomial

[UVA][Math] 12024 - Hats

[UVA][數學][繁瑣] 10367 - Equations

[UVA] 10177 - (2/3/4)-D Sqr/Rects/Cubes/Boxes?