高手进来解决个问题啊

[yaoyin0819]

输入：111332233
输出：31232223
输入：1111111111
输出：101

意思就是：连续3个1连续2个3连续2个2连续2个3 输出结果就是31232223
连续10个1 输出结果就是101

哪位高手能解决吗？

[paidfighting]

用字符串就行了

[yaoyin0819]

字符串我知道啊 问题是怎么做？ 我是新手~~

[fallinleave]

如果你的初始输入是input,

int* input;
int* output;

int nLen;

//Get the input;
....

nLen = length of input;

output = new int[2*nLen];

for(int i=0, j=0; i<nLen-1; i++)
{
output[j+1] = input[i];
int count = 0;
while(input[i] == input[i+1])
{
i++;
count++;
}
output[j] = count + 1;
j = j + 2;
}

[c_spark]

#include <stdio.h>
#include <string.h>

#define MAXSIZE 100

int main()
{
	int i,count;
	char str[MAXSIZE];
	while(scanf("%s",str) != EOF)
	{
		for(i = 1,count = 1; i < strlen(str); ++i)
		{
			if(str[i-1] == str[i])
			{
				++count;
				continue;
			}
			printf("%d%c",count,str[i-1]);
			count = 1;		
		}
		printf("%d%c\n",count,str[i-1]);
	}
	return 0;
}

这个看看行不?

[popyy0101]

#include <iostream>
#include <string>
using namespace std;
const int size = 100;
int main()
{
char s[size];
cin.getline(s,size);
int i = 0;
int count = 0;
int oldNum;
int num;
char c;
while(s[i] != 0){
c = s[i++];
num = atoi(&c);
if( i == 1 || num == oldNum){
count++;
}
else{
cout << count << oldNum;
count = 1;
}
oldNum = num;
}
cout << count << oldNum;
}
应该差不多吧

[劲草]

#include "stdio.h"
#include "stdlib.h"
#include "iostream.h"
void main()
{
char ch[20];
int i=0,j=1;
gets(ch);


while(ch[i]!='\0')
{
if(ch[i]>='0'&&ch[i]<='9')
{
if(ch[i+1]==ch[i])
{
j++;i++;
}
else
{
cout<<j<<ch[i];
j=1;i++;
}

}

else
cout<<"您的输入有误！"<<endl;

}


}

[rushman]

#include <stdio.h>

void main(void){
  char input[40];
  char * pc;
  char cc;
  int count;

  while(gets(input)){
	if(0 == *input)break;
    for(pc = input,cc = *input,count = 0;*pc != '\0';pc++){
      if(cc != *pc){
        printf("%d%c",count,cc);
        cc = *pc;
        count = 1;
        continue;
      }
      count++;
    }
    printf("%d%c\n",count,cc);
  }
}

[JYYCOM]

用堆栈应该也可以吧。

[shore1111]

提示：既然用到统计字符个数，最好用容器里的count() 函数喽
学了C++，最好用新的东西~~
字符的操作最好用容器来处理~~