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this is an old discussion but is there any suggestions about this sol in c
/**/
void repeat_char(char c , int co )
{
for (int i=0; i < co; i++ )
{
printf("%c", c);
}
}
int main(){
int t;
scanf("%d",&t);
int n[t];
for(int a0 = 0; a0 < t; a0++){
scanf("%d",&n[a0]);
}
for(int a0 = 0; a0 < t; a0++){
int num5 = n[a0] - n[a0] %3;
int num3 = n[a0] - num5;
switch(num3 % 5){
case 1: num3 = 10;break;
case 2: num3 = 5;break;
case 3: num3 = 15;break;
case 4: num3 = 10;break;
}
if(num3>n[a0]){
printf("-1\n");
continue;
}
num5 = n[a0] - num3;
if(num5>0)
repeat_char('5',num5);
if(num3>0)
repeat_char('3',num3);
printf("\n");
Sherlock and The Beast
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this is an old discussion but is there any suggestions about this sol in c /**/ void repeat_char(char c , int co ) { for (int i=0; i < co; i++ ) { printf("%c", c); } } int main(){ int t; scanf("%d",&t); int n[t]; for(int a0 = 0; a0 < t; a0++){ scanf("%d",&n[a0]); } for(int a0 = 0; a0 < t; a0++){
int num5 = n[a0] - n[a0] %3; int num3 = n[a0] - num5; switch(num3 % 5){ case 1: num3 = 10;break; case 2: num3 = 5;break; case 3: num3 = 15;break; case 4: num3 = 10;break; } if(num3>n[a0]){ printf("-1\n");
continue; } num5 = n[a0] - num3; if(num5>0) repeat_char('5',num5); if(num3>0) repeat_char('3',num3); printf("\n");
}