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| 1 | #include <stdio.h> | |
| 2 | #include <math.h> | |
| 3 | ||
| 4 | main(){
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| 5 | int n,i,a,p,somme,count; | |
| 6 | count=0; | |
| 7 | for( n=1; n<1000; n++ ){
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| 8 | somme=0; | |
| 9 | a=n; | |
| 10 | - | p=((int) ((log(n))/(log(2)))); |
| 10 | + | p=(int) ((log(n))/(log(2))); |
| 11 | for( i=1; i<=p; i++ ){
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| 12 | a=(int) a/2; | |
| 13 | somme+=a; | |
| 14 | } | |
| 15 | if( somme==n-1 ){
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| 16 | count++; | |
| 17 | printf("%d ", n);
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| 18 | } | |
| 19 | } | |
| 20 | printf("\nNombre d'entiers repondant au probleme : %d\n", count);
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| 21 | } |
