POJ 1416 Shredding Company (dfs)

题意：给定一个目标数n1，一个操作数n2，需要对操作数进行切割，并且各个和要小于目标数，而且最大。

由于操作数n2不大,dfs就可以实现，就是小细节很多，需要注意，对0的判断要加上。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <vector>
using namespace std;

const int BORDER = (1<<26)-1;
#define MAXN 1000000
#define INF 0x7ffffff
#define _clr(x,y) memset(x,y,sizeof(x))
#define _add(x) ((++x)&BORDER)
#define _in(x) scanf("%d",&x)
#define _out(x) printf("%d\n",x)
#define _min(m,v) (m)<(v)?(m):(v)
#define _max(m,v) (m)>(v)?(m):(v)

int n,target,arr[10],sum,mmax,tar_len,t_sum,n_index;
int pre[10],t_pre[10],cnt[MAXN],arr_sum[12];
int init()
{
    _clr(pre,0);
    _clr(cnt,0);
    _clr(arr,0);
    _clr(t_pre,0);
    mmax = -1;
    arr[0] = 0;
    target = 0;
    return 0;
}
int dfs(const int& ilen,const int& cur_pos,const int& index)
{
    int i,j,len,tmp;
    t_pre[index] = ilen;
    t_sum = 0;
    for(i = ilen-1; i >=0; --i)
        t_sum = t_sum*10 + arr[cur_pos-i];
    if(!index)
        arr_sum[index] = t_sum;
    else
        arr_sum[index] = arr_sum[index-1] + t_sum;
    tmp = arr_sum[index];
    if(    tmp > target)
        return 0;
    if(cur_pos == n)
    {
        if( tmp == mmax )
            cnt[tmp] = 1;
        else if(tmp > mmax)
        {
            mmax = tmp;
            n_index = index;
            memcpy(pre,t_pre,sizeof(pre));
        }
    }
    len = n-cur_pos;
    for(i = 1; i <= len; ++i)
        dfs(i,cur_pos+i,index+1);

    return 1;
}
int main()
{
    char str1[10],str2[10];
    int i,j,tmp;
    while(scanf("%s %s",str1,str2))
    {
        init();
        if(str1[0] == '0' && str2[0] == '0')
            break;
        tar_len = strlen(str1);
        for(i = 1; i <= tar_len; ++i)
            target = target*10 + str1[i-1]-'0';
        n = strlen(str2);
        for(i = 1; i <= n; ++i)
            arr[i] = str2[i-1] - '0';
        dfs(0,0,0);
        if(mmax == -1)
            printf("error\n");
        else if(cnt[mmax])
            printf("rejected\n");
        else
        {
            printf("%d",mmax);
            int cur = 1;
            for(i = 1;i<=n_index; ++i)
            {
                tmp = pre[i];
                printf(" ");
                for(j = 0; j < tmp; ++j)
                {
                    printf("%c",arr[cur]+'0');
                    ++cur;
                }
            }
            printf("\n");    
        }
    }
    return 0;
}