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

Background

Lack of rain in Spain is becoming a serious problem. A decrease in annual rainfall since the 1970s is said to have produced the present low reservoir levels. Although reservoirs across the country are running at an average capacity of less than 50 percent, in the worst affected areas such as Murcia, the capacity in many reservoirs is said to be less than 20 percent.

You are a scientist that must compute each month the capacity, in percentage, of water of a given reservoir.

The Problem

A reservoir is described by a list of coordinates (P[1], P[2], ..., P[N]) which define its cross section, and the width W of the reservoir, as shown in the figure below. The points that define the cross section always go from the left to the right, i.e., P[i].x <= P[i+1].x.

Cross section of the reservoir: Both dimensions and coordinates are given in metres.

We know the level of the reservoir at the beginning of the month (as a percentage). Also, we know the amount of water consumed due to evaporation and population use, and the amount of water falling on the reservoir because of rain (both measures are given in cubic metres). We suppose that water consumption only happens in the first half of the month, and the rain in the second half (so, in the same month, we can have a lack of water and then an excess of water).

The Input

The first line of the input contains an integer T (0 < T < 200), indicating the number of test cases. Test cases are separated by blank lines.

Each test case starts with a line containing the number N (2 < N < 10000) of coordinate pairs (x, y) of the reservoir, with x >= 0 and y <= 0. The following N lines for each test case contain the coordinates defining the reservoir's cross section (two integer numbers per line). The first and last coordinates always have y = 0. Coordinates are always in order from left to right. The following line contains only an integer W indicating the width of the reservoir.

Finally, each test case ends with a line containing three real numbers representing: the initial percentage of water in the reservoir, the amount of spent water (in cubic metres), and the amount of water falling over the reservoir (also in cubic metres), respectively.

The Output

For each test case, you must write a line with the text "Final percentage: X%", where X is an integer number between 0 and 100 indicating the final capacity (as a percentage) of the reservoir; you have to take the integer part of the result (rounding down).

If the amount of water consumed is greater than the initial capacity of the reservoir, you must write the string "Lack of water. " before the previous text (observe that it ends with a blank space).

If the final amount of water exceeds the potential capacity of the reservoir, you must write "Excess of water. " (observe that it ends with a blank space) before the final percentage and, if it appears, after the "lack of water" text.

Note that in some cases the three texts may appear. Remember: first there is a water consumption and then water recovering by the rain.

Sample Input

3

4
0 0
0 -2
2 -2
2 0
1
50.0 2.0 1.0

9
0 0
9 -2
11 -1
16 -3
22 -3
23 0
28 -2
28 -4
36 0
2
84.2 178.24 991.00

3
0 0
1 -1
2 0
100
0.01 23.00 49.99

Sample Output

Final percentage: 25%
Lack of water. Excess of water. Final percentage: 100%
Lack of water. Final percentage: 49%

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仔細看題目其實也還好，還以為給定水量，二分搜水位高度。
中文翻譯如下：

http://unfortunatedog.blogspot.tw/2013/07/11447-reservoir-logs.html

計算多邊形面積即可。

#include <stdio.h>
#include <math.h>
double x[10005], y[10005];
int main() {
    int testcase, n;
    int i, j;
    scanf("%d", &testcase);
    double w, percent, consume, rain;
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%lf %lf", &x[i], &y[i]);
        x[n] = x[0], y[n] = y[0];
        double area = 0;
        for(i = 0; i < n; i++)
            area += x[i]*y[i+1]-y[i]*x[i+1];
        area = fabs(area)/2;
        scanf("%lf", &w);
        scanf("%lf %lf %lf", &percent, &consume, &rain);
        double V = area * w * (percent/100.0);
        V -= consume;
        if(V < 0) {
            printf("Lack of water. ");
            V = 0;
        }
        V += rain;
        if(V > area * w) {
            printf("Excess of water. ");
            V = area * w;
        }
        printf("Final percentage: %d%%\n", (int)(V/(area*w)*100));
    }
    return 0;
}