解题思路:
输入一个数N,循环判断,从i=2到N的每一个数;
如果是完数,则输出;
判断方法,定义j为因子(1到N-1),若i%j==0,则是它的因子;
判断期间把所有因子的和(sum)求出来;
最后和(sum)等于i则是完数;
注意事项:
每次判断完,一个数i后,要把sum置0;
按输出规则输出;
参考代码:
#include <stdio.h> int main() { long double N, sum = 0; scanf( "%Lf", &N ); for ( int i = 2; i <= N; i++ ) { for ( int j = 1; j <i; j++ ) { if ( i % j == 0 ) { sum = sum + j; } } if ( sum == i ) { printf( "%d its factors are ", i ); for ( int k = 1; k < i; k++ ) { if ( i % k == 0 ) { printf( "%d ", k ); } } printf( "\n" ); } sum = 0; } return(0); }
上述代码已被淘汰,现在提交会时间超限;
根据一个数的最大因子(除去他本身)不会大于它的一半,改进得到下面代码(注意:一个数 n 如果是合数,那么它的所有的因子不超过sqrt(n)是错误的;比如16的一个因子8,大于4;)
改进代码:
#include <stdio.h> int main() { long double N, sum = 0; scanf( "%Lf", &N ); for ( int i = 2; i <= N; i++ ) { for ( int j = 1; j <=i/2; j++ ) { if ( i % j == 0 ) { sum = sum + j; } } if ( sum == i ) { printf( "%d its factors are ", i ); for ( int k = 1; k <=i/2; k++ ) { if ( i % k == 0 ) { printf( "%d ", k ); } } printf( "\n" ); } sum = 0; } return(0); }
改进代码2:(观察前面12 个完数,个位数都是6 或者8,所以个位不是6 或8的就不判断了,并且在判断过程中,直接记录它的因子)
#include <stdio.h> int main() { long double N, sum = 0; int factor[100], x = 0; scanf( "%Lf", &N ); for ( int i = 6; i <= N; i++ ) { if ( i % 10 == 6 || i % 10 == 8 ) { for ( int j = 1; j <= i / 2; j++ ) { if ( i % j == 0 ) { factor[x] = j; x++; sum = sum + j; } } if ( sum == i ) { printf( "%d its factors are ", i ); for ( int k = 0; k < x; k++ ) { printf( "%d ", factor[k] ); } printf( "\n" ); } sum = 0; x = 0; } } return(0); }
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#include <stdio.h> int main(){ int N; scanf("%d",&N); int number[N]; int i,j,sum=0; for(i=0;i<N;i++){ number[i]=i+4; sum=0; for(j=1;j<number[i];j++){ if(number[i]%j==0){ sum+=j; if(number[i]==sum){ printf("%d\n",number[i]); } } } } return 0; } 各位大佬,能不能帮忙看看,当我键入N=1000的时候,会多一个24出来,其他的结果和题目给的例子是符合的,想知道是哪里出了问题
#include<stdio.h> int main() { int N; scanf("%d",&N); int m,n,p; for(m=1;m<=N;m++) { int n=0; for(p=1;p<m;p++) { if(m%p==0) n=p+n; } if(n==m) {printf("%d its factors are ",m); for(p=1;p<m;p++) { if(m%p==0) printf("%d ",p); } printf("\n"); } } return 0; } 我的最后没用数组
老哥们看看我哪错了,为啥提交总是答案错误啊 #include<stdio.h> int main() { long double sum=0,N; int i,k,j,factor[100],x=0; scanf("%Lf",&N); for(i=6;i<=N;i++) { if(i%10==6||i%10==8) { for(j=1;j<=i/2;j++) { if(i%j==0) { factor[x]=j; x++; sum=sum+j; } } if(sum==i) { printf("%d its factors are",i); for(k=0;k<x;k++) { printf("%d",factor[k]); } printf("\n"); } sum=0;x=0; } } return 0; }
#include<stdio.h> int main() { int N; int i,j,a; scanf("%d",&N); for(i=1;i<=N;i++) { a=0; for(j=1;j<i;j++) { if(i%j==0) a+=j; } if(i==a) { printf("%d its factors are ",i); for(j=1;j<i;j++) { if(i%j==0) printf("%d ",j); } printf("\n"); } } return 0; }
Manchester的第三种比较厉害 我也给出我的吧 #include <stdio.h> #include <stdlib.h> int main() { int N,i,j,k,a[1001],sum; scanf("%d",&N); for(i=2; i<=N; i++) { sum=0,k=1; for(j=1; j<=i/2; j++) { if(i%j==0) { a[k]=j; sum+=a[k]; k++; } } if(sum==i) { printf("%d its factors are",i); for(j=1; j<k; j++) printf(" %d",a[j]); printf("\n"); } } return 0; }
#include <stdio.h> int main() { int n; scanf("%d",&n); int i,j,m,sum; for( m=2;m<=n;m++ ){ sum = 0; for( i=1;i<m;i++ ){ if( m%i == 0 ){ sum += i; } } if( sum == m ){ printf("%d its factors are ",m); for( j=1;j<m;j++ ){ if( m%j==0 ){ printf("%d ",j); } } printf("\n"); } } return 0; }
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