Question

我正在尝试制作一个显示飞机到达和离开时间表的程序，然后要求用户在C中输入一个时间。然后该程序将找到最接近该时间的到达时间，由用户。问题是它没有按预期方式工作，并且给我错误的到达时间，甚至还没有接近输入时间。我是C编程的新手，请原谅我凌乱的编程/逻辑。在这里：

#include <stdio.h>

int main(void) {

int time0, time1, time2, time3, time4, time5, time6, time7, time8, t, m, i1 
= 1, i2 = 3, i3 =4, i4 =5, i5=6, i6=7, i7=8, i8=9, u1=10, u2=20, u3=30, 
u4=40, u5=50, u6=59, u7=25, u8=45, y1=13, y2=14, y3=15, y4=16, y5=17, y6=18, 
y7=19, y8=20, h1=17, h2=18, h3=19, h4=30, h5=40, h6=40, h7=35, h8=7, g1, g2, 
g3, g4, g5, g6, g7, g8, c, v;

/* The timetable, in which the majority of the variables are used */

printf("Arrival\tDeparture\n");
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i1, u1, y1, h1);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i2, u2, y2, h2);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i3, u3, y3, h3);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i4, u4, y4, h4);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i5, u5, y5, h5);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i6, u6, y6, h6);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i7, u7, y7, h7);
printf("%3.2d:%.2d\t%3.2d:%.2d\n", i8, u8, y8, h8);
printf("Enter your time for departure: ");
scanf("%d:%d", &t, &m);

/*Breaks down the entered time and the times from the timetable in minutes 
since midnight for comparing, later in the program */

time0= 60*t +m;  
time1= 60*i1 +u1;  
time2= 60*i2 +u2;  
time3= 60*i3 +u3;  
time4= 60*i4 +u4;  
time5= 60+i5 +u5;  
time6= 60+i6 +u6;  
time7= 60+i7 +u7;  
time8= 60+i8 +u8;  

/* subtracts the different arrival times from the entered time if the 
arrival times are bigger, else the entered time is subtracted and saves the 
difference in a variable for later use */

if(time0 > time1) {
    g1 = time0 -time1;
} else {
    g1 = time1 -time0;
}  
if(time0 > time2) {
    g2 = time0-time2;
} else {
    g2 = time2 - time0;
}  
if(time0 > time3) {
    g3 = time0- time3;
} else {
    g3 = time3 -time0;
}  
if(time0 > time4) {
    g4 = time0 - time4;
} else {
    g4 = time4 - time0;
}  
if(time0 > time5) {
    g5=time0 -time5;
} else {
    g5= time5 - time0;
}  
if(time0 > time6) {
    g6 = time0-time6;
} else {
    g6 =time6-time0;
}  
if(time0 > time7) {
    g7 = time0-time7;
} else {
    g7 =time7-time0;
}  
if(time0 > time8) {
    g8 = time0-time8;
} else {
    g8 =time8-time0;
}  
/* here, the program should compare all the differences, to check which one 
is the smallest, and then save the smallest hour(s) and minute(s) in the v 
and c variables  */  

v = i1;
c = u1;

if( g2 < g1 )  {
    v = i2;
    c = u2;
}  
if( g3 < g2)  {
    v = i3;
    c = u3;
}  
if( g4 < g3) {
    v = i4;
    c = u4;
}  
if( g5 < g4) {
    v = i5;
    c = u5;
}  
if( g6 < g5) {
    v = i6;
    c = u6;
}  
if( g7 < g6) {
    v = i7;
    c = u7;
}  
if( g8 < g7) {
    v = i8;
    c = u8;
}  
/* The time with the smallest difference to the entered time should be 
printed*/  
printf("The closest time is %d:%d", v, c);
}

但是它打印了错误的时间，我找不到错误。

Answer 1

我认为问题在于代码中的逻辑：

v = i1;
c = u2;

if( g2 < g1 )  {
    v = i2;
    c = u2;
}  
if( g3 < g2)  {
    v = i3;
    c = u3;
}  
if( g4 < g3) {
    v = i4;
    c = u4;
}

考虑案例g1 < g3 < g2。在这里您要选择g1，但实际上会发生什么？

1）首先，您假设g1是最小的。（精细）

2）然后，您比较g2和g1，但继续将g1最小（精细）

3）然后，您比较g2和g3并将g3选择为最小！（错误）

问题在于，您永远不会将g3与g1进行比较，因此即使实际上g3大于g1，也选择v = i1; c = u2; smallest = g1; if( g2 < smallest ) { v = i2; c = u2; smallest = g2; } if( g3 < smallest) { v = i3; c = u3; smallest = g3; } if( g4 < smallest) { v = i4; c = u4; smallest = g4; }。那不是你想要的。

您需要的是一个额外的变量，该变量保持当前的最小值。喜欢：

magnify

顺便说一句：我建议您尽快学习数组，因为这将简化您的代码

到达/离开程序

1 个答案: