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                [TOC] ### 題目描述 如果你是哈利·波特迷,你會知道魔法世界有它自己的貨幣系統 —— 就如海格告訴哈利的:“十七個銀西可(Sickle)兌一個加隆(Galleon),二十九個納特(Knut)兌一個西可,很容易。”現在,給定哈利應付的價錢P和他實付的錢A,你的任務是寫一個程序來計算他應該被找的零錢。 輸入格式: 輸入在1行中分別給出P和A,格式為“Galleon.Sickle.Knut”,其間用1個空格分隔。這里Galleon是[0, 107]區間內的整數,Sickle是[0, 17)區間內的整數,Knut是[0, 29)區間內的整數。 輸出格式: 在一行中用與輸入同樣的格式輸出哈利應該被找的零錢。如果他沒帶夠錢,那么輸出的應該是負數。 輸入樣例1: 10.16.27 14.1.28 輸出樣例1: 3.2.1 輸入樣例2: 14.1.28 10.16.27 輸出樣例2: -3.2.1 代碼: ``` #include <stdio.h> int main() { //不能整型讀入數據,因為計算會把小數點讓成浮點數的一部分,讀入浮點數后再存入整型變量 char str[50]; gets(str); int g1 = 0, s1 = 0, k1 = 0; int g2 = 0, s2 = 0, k2 = 0; int i = 0; //字符串讀入后轉化成整型變量 while(str[i] != '.') g1 = 10 * g1 + str[i++] - '0'; ++i; while(str[i] != '.') s1 = 10 * s1 + str[i++] - '0'; ++i; while(str[i] != ' ') k1 = 10 * k1 + str[i++] - '0'; ++i; while(str[i] != '.') g2 = 10 * g2 + str[i++] - '0'; ++i; while(str[i] != '.') s2 = 10 * s2 + str[i++] - '0'; ++i; while(str[i] != '\0') k2 = 10 * k2 + str[i++] - '0'; int g = g2 - g1, s = s2 - s1, k = k2 - k1; if(k < 0) { //計算找零 --s; k += 29; } if(s < 0) { --g; s += 17; } //錢不夠時不能直接前面添個符號,那樣后兩個數字仍然表示找回的錢,而不是欠的錢 if(g < 0) { g = g1 - g2; s = s1 - s2; k = k1 - k2; if(k < 0) { --s; k += 29; } if(s < 0) { --g; s += 17; } g = -g; } printf("%d.%d.%d", g, s, k); return 0; } ```
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