解题思路:
设n=5；则sum=a+aa+aaa+aaaa+aaaaa;
分析：
1.n个数相加，考虑到进位如99999+9999，需要长度为n+1的数组存放结果；
2.如图，个位结果为n-0个a的和，十位结果为n-1个a的和...万位为n-4个a的和；
for ( int i = n; i > 0; i-- ) sum[i] += i * a;
3.求出每位的和后，处理进位：从个位数开始，每一位先向前面一位进位，然后再求这一位对应数值（图中先蓝色，后绿色）
for ( int i = n; i > 0; i-- )
{
sum[i - 1] += (sum[i] / 10);
sum[i] %= 10;
}4.输出结果：如果sum[0]==0的话，最高位没有收到进位，输出sum[1]到sum[n],否则输出sum[0]到sum[n];
if ( sum[0] == 0 )
{
for ( int i = 1; i <= n; i++ )
printf( "%d", sum[i] );
}else
for ( int i = 0; i <= n; i++ )
printf( "%d", sum[i] );注意事项:
开辟数组长度为n+1；最高位数下标为0，个位数下标为n；
参考代码: a为正
#include <stdio.h>
#include <malloc.h>
void format( int *sum, int n );
void function( int *sum, int a, int n );
/*===================================================*/
int main()
{
int a, n;
int *sum;
scanf( "%d%d", &a, &n );
sum = (int *) malloc( (n + 1) * sizeof(int) );
format( sum, n );
function( sum, a, n );
return(0);
}
/*===================================================*/
void format( int *sum, int n )
{
for ( int i = 0; i <= n; i++ )
sum[i] = 0;
}
/*===================================================*/
void function( int *sum, int a, int n )
{
for ( int i = n; i > 0; i-- )
sum[i] += i * a;
/*------------------------------------*/
for ( int i = n; i > 0; i-- )
{
sum[i - 1] += (sum[i] / 10);
sum[i] %= 10;
}
/*------------------------------------*/
if ( sum[0] == 0 )
{
for ( int i = 1; i <= n; i++ )
printf( "%d", sum[i] );
}else
for ( int i = 0; i <= n; i++ )
printf( "%d", sum[i] );
return ;
}等哪天上面代码不正确了：说明测试数据有a为负数，那么用下面代码正负a都对，a为负，会加负号；
#include <stdio.h>
#include <malloc.h>
#include <math.h>
void format( int *sum, int n );
void function( int *sum, int a, int n );
/*===================================================*/
int main()
{
int a, n;
int *sum;
scanf( "%d%d", &a, &n );
sum = (int *) malloc( (n + 1) * sizeof(int) );
format( sum, n );
function( sum, a, n );
return(0);
}
/*===================================================*/
void format( int *sum, int n )
{
for ( int i = 0; i <= n; i++ )
sum[i] = 0;
}
/*---------------------------------------------------*/
void function( int *sum, int a, int n )
{
for ( int i = n; i > 0; i-- )
sum[i] += i * fabs( a );
/*------------------------------------*/
for ( int i = n; i > 0; i-- )
{
sum[i - 1] += (sum[i] / 10);
sum[i] %= 10;
}
/*------------------------------------*/
if ( sum[0] == 0 )
{
if ( a < 0 )
printf( "-" );
for ( int i = 1; i <= n; i++ )
printf( "%d", sum[i] );
}else{
if ( a < 0 )
printf( "-" );
for ( int i = 0; i <= n; i++ )
printf( "%d", sum[i] );
}
return;
}别忘点赞哦-.-
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#include<stdio.h> #include<math.h> int main() { int a,n,s=0,i,amount; double b,c; scanf("%d %d",&a,&n); c=b=a; for(i=1;i<=n;i++) { amount=c*pow(10,(i-1)); c=c+b*pow(10,-i); s=s+amount; } printf("%d\n",s); } 大佬求解，为什么我这个代码是错的#include<bits/stdc++.h> using namespace std; int main() { int a,n; long long s,d; s=0; d=0; cin>>a>>n; for(int i=0;i<n;i++) { d=d+a*pow(10,i); s+=d; } cout<<s; }#include<stdio.h> int main() { int a, n, i, S; printf("please input a and n:"); scanf("%d %d", &a, &n); if(a <= 9 && n >= 1) { S = a; for(i = 2; i <= n; i++) { a = (10 * a) + (a % 10); S += a; printf("a = %d\n",a); } printf("S = %d\n", S); } return 0; }1 #include<stdio.h> 2 3 4 int main() 5 { 6 int a, n, i, S; 7 printf("please input a and n:"); 8 scanf("%d %d", &a, &n); 9 if(a <= 9 && n >= 1) 10 { 11 S = a; 12 for(i = 2; i <= n; i++) 13 { 14 a = (10 * a) + (a % 10); 15 S += a; 16 printf("a = %d\n",a); 17 } 18 19 printf("S = %d\n", S); 20 21 } 22 23 } ~