Created
May 24, 2016 23:36
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/** | |
Los puntos claves son usar gcd para reducir al maximo la fraccion sum/n | |
Para impresión se puede usar este tip: | |
c=cifras(n) | |
char b[50]="% d\n"; | |
b[1]=c+'0';//convertir int to char | |
printf(b,abs(sum)) | |
Asi se fuerza a ocupar c espacios la impresion del numero sum | |
**/ | |
#include <iostream> | |
#include <cmath> | |
#include <algorithm> | |
#include <queue> | |
#include <stack> | |
#include <sstream> | |
#include <map> | |
#include <set> | |
#include <queue> | |
#include <cstdio> | |
using namespace std; | |
int cifras(int x){ | |
int r=0; | |
x=abs(x); | |
for(int i=1;x>=i;i*=10){ | |
r++; | |
} | |
return r; | |
} | |
int n,sum,e; | |
int main(){ | |
scanf("%d",&n); | |
e=1; | |
while(n){ | |
sum=0; | |
for(int i=0;i<n;i++){ | |
int aux; | |
scanf("%d",&aux); | |
sum+=aux; | |
} | |
printf("Case %d:\n",e); | |
if(abs(sum)%n==0){ | |
if(sum<0) | |
printf("- "); | |
printf("%d\n",abs(sum)/n); | |
} | |
else if(abs(sum)<n){ | |
int gcd=__gcd(abs(sum),n); | |
sum/=gcd; | |
n/=gcd; | |
int c=cifras(n); | |
char b[50]="% d\n"; | |
b[1]=c+'0';//convertir int to char | |
if(sum<0) | |
printf(" "); | |
printf(b,abs(sum)); | |
if(sum<0) | |
printf("- "); | |
for(int i=0;i<c;i++) | |
printf("-"); | |
printf("\n"); | |
if(sum<0) | |
printf(" "); | |
printf("%d\n",n); | |
} | |
else{ | |
int ent=abs(sum/n); | |
sum%=n; | |
int gcd=__gcd(abs(sum),n); | |
sum/=gcd; | |
n/=gcd; | |
int c=cifras(n); | |
char b[50]="% d\n"; | |
b[1]=c+'0';//convertir int to char | |
int c2=cifras(ent); | |
if(sum<0) | |
printf(" "); | |
for(int i=0;i<c2;i++) | |
printf(" "); | |
printf(b,abs(sum)); | |
if(sum<0) | |
printf("- "); | |
printf("%d",ent); | |
for(int i=0;i<c;i++) | |
printf("-"); | |
printf("\n"); | |
if(sum<0) | |
printf(" "); | |
for(int i=0;i<c2;i++) | |
printf(" "); | |
printf("%d\n",n); | |
} | |
scanf("%d",&n); | |
e++; | |
} | |
return 0; | |
} |
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