卖苹果股票（1st In 1st Out）

任何人都可以帮助我使用我的代码，当我卖掉我的苹果时，我有一个问题，它会减去我所有的股票，有时当我的投入超过股票时这个数字会变为负数

int main(void){ int choice = 0; int days = 1, i, buyApple; int stocks[99] = {0}; for (;;){ clrscr(); printf("Day %d\n", days); printf("1. harvest\n"); printf("2. View Stocks\n"); printf("3. Sell\n"); printf("\nchoice: "); scanf("%d", &choice); if (choice == 1){ clrscr(); printf("Input No. of Apple harvested: "); scanf("%d", &stocks[days]); days++; } if (choice == 2){ clrscr(); printf("Day Stocks\n"); for (i = 1; i < days; i++){ printf("%2d %4d\n", i, stocks[i]); } getch(); } if(choice == 3){ printf("Input No. of Appple to be sold: "); scanf("%d", &buyApple); for (i = 0; i  buyApple) printf("Out of Stocks!"); getch(); } } } 

期待输出：

比如我的股票就是这个

  Day Stocks 1 100 2 50 3 50 4 180 5 200 

如果我输入的话，要售出的苹果是200

我的股票会变成这样

 Day Stocks 1 0 2 0 3 0 4 180 5 200 

排在第1位

当我的投入超过股票时，它会说"Out of Stocks !" 它不会继续减少我的股票

Answer 1

问题评论中的第三选择问题，正确的代码应该是这样的（它是c＃）：

 if (choice == 3) { printf("Input No. of Appple to be sold: "); scanf("%d", &buyApple); for (i = 1; i < sizeof(stocks); i++) //Looping on all stocks you have { if (buyApple <= stocks[i]) //if amount of apple less than stock apples ,then remove them from stock { stocks[i] = stocks[i] - buyApple; buyApple = 0; break; } else //if amount of apple is bigger than stock apples { if ((sizeof(stocks) - 1) == i){ //if it's the Last stock,then there is no apples in stock printf("Out of Stocks!"); } else //take amount of apples from current stock { buyApple = buyApple - stocks[i]; stocks[i] = 0; } } } getch(); }