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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