C语言，一道很有意思的传送带问题，望帮助。

Question

现有一玩具制作车间，一个玩具的正常产出需要经过五个工序，车间一旦启动，不可再中途终止。其中，传送带每次只能运行一个玩具，即需将前一个玩具送达之后才能传送第二个。玩具组装部分可同时组装8个玩具，玩具刷漆部分可同时工作4个玩具。每一类玩具在组装、刷漆部分的历时不尽相同。玩具组装时长50、刷漆时长22，传送带1时长8，传送dai2时长5，传送d3时长5。请问产生180个玩具需要多少时间

Answer 1

#include <stdio.h>
 
#define ASSEMBLY_TIME 50
#define PAINTING_TIME 22
#define CONVEYOR1_TIME 8
#define CONVEYOR2_TIME 5
#define CONVEYOR3_TIME 5
 
void processProduct(int productNumber, int startTime) {
    int assemblyStartTime = startTime;
    int assemblyEndTime = assemblyStartTime + ASSEMBLY_TIME;
    printf("产品%d在%d秒进入组装部分，在%d秒离开组装部分，", productNumber, assemblyStartTime, assemblyEndTime);
 
    int paintingStartTime = assemblyEndTime;
    int paintingEndTime = paintingStartTime + PAINTING_TIME;
    printf("在%d秒进入刷漆部分，在%d秒离开刷漆部分，", paintingStartTime, paintingEndTime);

    int conveyor1StartTime = paintingEndTime;
    int conveyor1EndTime = conveyor1StartTime + CONVEYOR1_TIME;
    printf("在%d进入传送带1，在%d秒离开传送带1，", conveyor1StartTime, conveyor1EndTime);
 
    int conveyor2StartTime = conveyor1EndTime;
    int conveyor2EndTime = conveyor2StartTime + CONVEYOR2_TIME;
    printf("在%d秒进入传送带2，在%d秒离开传送带2，", conveyor2StartTime, conveyor2EndTime);
 
    int conveyor3StartTime = conveyor2EndTime;
    int conveyor3EndTime = conveyor3StartTime + CONVEYOR3_TIME;
    printf("在%d秒进入传送带3，在%d秒离开传送带3\n", conveyor3StartTime, conveyor3EndTime);
}
 
int main() {
    int productCount = 180;
 
    int conveyor1Time = 0;
    int conveyor2Time = 0;
    int conveyor3Time = 0;
 
    for (int i = 1; i <= productCount; i++) {
        int startTime = conveyor1Time > conveyor2Time ? conveyor1Time : conveyor2Time;
        startTime = startTime > conveyor3Time ? startTime : conveyor3Time;
 
        processProduct(i, startTime);
 
        conveyor1Time = startTime + CONVEYOR1_TIME;
        conveyor2Time = startTime + CONVEYOR2_TIME;
        conveyor3Time = startTime + CONVEYOR3_TIME;
    }
 
    return 0;
}