1 solutions
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0
C :
#include<stdio.h> #include<string.h> int main() { int n,i,j,l[10]; char x[10][51]; scanf("%d",&n); getchar(); for(i=0;i<n;i++) { gets(x[i]); l[i]=strlen(x[i]); } for(i=0;i<n;i++) { for(j=0;j<l[i];j++) { if(x[i][j]=='0') x[i][j]='1'; else x[i][j]='0'; } x[i][j+1]='\0'; } for(i=0;i<n;i++) { printf("%s",x[i]); if(i!=n-1) printf("\n"); } }
- 1
Information
- ID
- 17274
- Time
- 1000ms
- Memory
- 2MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By