Have a class C, geometry. A freshman for help-CodePudding

7 to 18 have class geometry (10)
Xiao Ming saw a course on the schedule of weekly in class is marked as 1, 8, 12-17, 3-6, would you please write a program that will input from the input devices such as the similar weekly said string, converted into standard weekly said,

Input format:
A digital said a week, the middle of the two Numbers - Numbers separated said one week time interval, a - b said range [a, b], procedures to ensure a
The output format:
From small to large, the order of the output with weeks of the course by commas between weeks and weeks, finally no comma,

Input the sample:
3-5,8,13-15,17,10-12
13 to 18, 5-9
3-13, 13-17
The output sample:
3,4,5,8,10,11,12,13,14,15,17
5,6,7,8,9,13,14,15,16,17,18
3,4,5,6,7,8,9,10,11,12,13,14,15,16,17

CodePudding user response:

//input format: 
//a digital said a week, the middle of the two Numbers - Numbers separated said one week time interval, a - b said range [a, b], procedures to ensure a//
//output format: 
//from the order of the output course weekly, comma-separated between weeks and weeks, finally no comma, 
//
//input sample: 
//3-5,8,13-15,17,10-12 
//13 to 18, 5-9 
//3-13, 13 to 17 
//output sample: 
//3,4,5,8,10,11,12,13,14,15,17 
//5,6,7,8,9,13,14,15,16,17,18 
//3,4,5,6,7,8,9,10,11,12,13,14,15,16,17 
#include  
#include  
#include  
Const int maxln=100; 
Char a, [maxln] [21]. 
Char ln [maxln] [80]; 
char *p; 
Int N, I, j, k, u, N, f, s, t, 
Int main () {
N=0; 
While (1) {
If (NULL==the fgets (ln [N], 80, stdin)) break; 
If (=='\ n' ln [n] [0]) break; 
N++; 
If (N>=maxln) break; 
} 

For (I=0; i
For (I=0; iP=& amp; Ln [I] [0]; 
S=1; 
While (1) {
K=sscanf (p, "% d % n, & amp; U, & amp; n); 
//printf (" k, u, n=% d, % d, % d \ n ", k, u, n); 
If (1)==k {
If (1==s) {f=u; T=u; S=2; } 
Else if (uThe else {t=u; S=1; } 
P +=n; 
{} else if (0==k) 
//printf (" skip char [c] % \ n ", p [0]). 
If (', '==p [0]) {
//printf (" f - t: % d, % d \ n ", f, t); 
For (j=f; J<=t; J++) a [I] [j]=1; 
S=1; 
} 
P++; 
} else {//EOF==k 
EOF f - t://printf (" % d, % d \ n ", f, t); 
For (j=f; J<=t; J++) a [I] [j]=1; 
break; 
} 
} 
N=0; 
For (j=1; J<=20; J++) 
If (1==a [I] [j]) {
If (0==n) {
Printf (" % d ", j); 
N++; 
} else {
Printf (" % d ", j); 
} 
} 
printf("\n"); 
} 
return 0; 
}