求解程序连续输出问题
程序连续运行,然后按某个键,退出运行窗口。怎样实现
#include <stdio.h> #include <conio.h> //要求用户输入股票数量和单价,采用某种比例计算经理人佣金,并计算竞争对手的佣金 //(小于2000 33美元+0.03)(大于等于2000 33美元+0.02) int main() { int quantity; float price, money, commission, commission1; printf("请输入股票数量: "); scanf("%d", &quantity); printf("请输入股票单价: "); scanf("%f", &price); money = quantity * price; if (money < 2500.00f) commission = 30.00f + 0.017f * money; else if (commission < 6250.00f) commission = 56.00f + 0.0066f * money; else if (money < 20000.00f) commission = 76.00f + 0.0034f * money; else if (money < 50000.00f) commission = 100.00f + 0.0022f * money; else if (money < 500000.00f) commission = 155.00f + 0.0011f * money; else commission = 255.00f + 0.0003f * money; if (commission < 39.00f) commission = 39.00f; printf("经理人的佣金: $%.2f\n", commission); if (quantity < 2000.00f) commission1 = 33.00f + quantity * .03f; else commission1 = 33.00f + quantity * .02f; printf("经理人竞争对手的佣金: $%.2f\n", commission1); printf("按任意键退出..");//亮点在这里。。。。。。。。。。。 getch(); return 0; }
int main(void) { char code; struct part inventory[MAX_PARTS]; int num_parts = 0; for (;;) { printf("Enter operation code: "); scanf(" %c", &code); while (getchar() != '\n') ; switch (code) { case 'i': insert(inventory, &num_parts); break; case 's': search(inventory, num_parts); break; case 'u': update(inventory, num_parts); break; case 'p': print(inventory, num_parts); break; case 'q': return 0; default: printf("Illegal code\n"); } printf("\n"); } }