poj 1416 Shredding Company （应该是用dfs吧，但是想了一下午不知道怎么递归来做，所以暴力做了）

Shredding Company

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5381		Accepted: 3023

Description

You have just been put in charge of developing a new shredder(碎纸机) for the Shredding Company Although a "normal" shredder would just
shred sheets of paper into little pieces so that the contents would become unreadable(不值一读的), this new shredder needs to have the
following unusual basic characteristics(特征).

1.The shredder takes as input(输入) a target number and a sheet of paper with a number written on it.

2.It shreds (or cuts) the sheet into pieces each of which has one or more digits(数字) on it.

3.The sum of the numbers written on each piece is the closest possible number to the target number, without going over it.

For example, suppose that the target number is 50, and the sheet of paper has the number 12346. The shredder(碎纸机) would cut the sheet
into four pieces, where one piece has 1, another has 2, the third has 34, and the fourth has 6. This is because their sum 43 (= 1 + 2 + 34 + 6) is closest to the target number 50 of all possible combinations(结合) without
going over 50. For example, a combination where the pieces are 1, 23, 4, and 6 is not valid, because the sum of this combination 34 (= 1 + 23 + 4 + 6) is less than the above combination's 43. The combination of 12, 34, and 6 is not valid either, because the
sum 52 (= 12 + 34 + 6) is greater than the target number of 50.



Figure 1. Shredding a sheet of paper having the number 12346 when the target number is 50

There are also three special rules :

1.If the target number is the same as the number on the sheet of paper, then the paper is not cut.

For example, if the target number is 100 and the number on the sheet of paper is also 100, then

the paper is not cut.

2.If it is not possible to make any combination whose sum is less than or equal to the target number, then error is printed on a display. For example, if the target number is 1 and the number on the sheet of paper is 123, it is not possible to make any valid
combination, as the combination with the smallest possible sum is 1, 2, 3. The sum for this combination is 6, which is greater than the target number, and thus error is printed.

3.If there is more than one possible combination where the sum is closest to the target number without going over it, then rejected is printed on a display. For example, if the target number is 15, and the number on the sheet of paper is 111, then there are
two possible combinations with the highest possible sum of 12: (a) 1 and 11 and (b) 11 and 1; thus rejected is printed. In order to develop such a shredder, you have decided to first make a simple program that would simulate(模仿的) the
above characteristics(特征) and rules. Given two numbers, where the first is the target number and the second is the number on the sheet
of paper to be shredded, you need to figure out how the shredder should "cut up" the second number.

Input

The input(投入) consists of several test cases, each on one line, as follows :

tl num1

t2 num2

...

tn numn

0 0

Each test case consists of the following two positive(积极的) integers(整数),
which are separated by one space : (1) the first integer (ti above) is the target number, (2) the second integer (numi above) is the number that is on the paper to be shredded.

Neither integers may have a 0 as the first digit(数字), e.g., 123 is allowed but 0123 is not. You may assume(承担) that
both integers are at most 6 digits in length. A line consisting of two zeros signals the end of the input.

Output

For each test case in the input, the corresponding output(输出) takes one of the following three types :

sum part1 part2 ...

rejected

error

In the first type, partj and sum have the following meaning :

1.Each partj is a number on one piece of shredded(切碎的) paper. The order of partj corresponds to the order of the original digits(数字)on
the sheet of paper.

2.sum is the sum of the numbers after being shredded, i.e., sum = part1 + part2 +...

Each number should be separated by one space.

The message error is printed if it is not possible to make any combination(结合), and rejected if there is

more than one possible combination.

No extra characters including spaces are allowed at the beginning of each line, nor at the end of each line.

Sample Input

50 12346
376 144139
927438 927438
18 3312
9 3142
25 1299
111 33333
103 862150
6 1104
0 0

Sample Output

43 1 2 34 6
283 144 139
927438 927438
18 3 3 12
error
21 1 2 9 9
rejected
103 86 2 15 0
rejected

Source

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>

using namespace std;
int a[10];
int b[10];
int q[1000000];
int n,m,ii;
int main()
{
while(~scanf("%d%d",&n,&m))
{
ii=0;
if(n==0&&m==0)break;
if(n==m)
{
p
4000
rintf("%d %d\n",n,m);
continue;
}
while(m)  //先把m分解并储存到a数组里边
{
++ii;
a[ii]=m%10;
m=m/10;

}

for(int i=1,j= ii; i<=j; i++,j--)//  交换顺序
{
int t=a[i];
a[i]=a[j];
a[j]=t;
}
int sum=0;
for(int i=1; i<=ii; i++)//判断将各个数都分开是否满足小于n,这时是最小和，如果不                                                满足，则输出error;
{
sum+=a[i];
b[i]=a[i];

}
if(sum>n)printf("error\n");
else
{
int su;
int ko=ii;
int mi;
int mx;
memset(q,0,sizeof(q));
for(int i=1; i<=ii; i++)//暴力枚举所有可能出现的情况
{
int x1=0;
mi=0;
mx=0;
if(i==ii)
{
for(int si=1; si<=ii; si++)
{
if(si<=ii)             //       只存在一个数
{
x1=x1*10+a[si];
}
}
su=x1;
if(su>=sum&&su<=n)
{
b[1]=x1;
ko=1;
sum=su;

}
q[su]++;
continue;
}
int jj=0;
while(jj<i)
{
mi=mi*10+a[++jj];
}
mx=mi;
if(mx>n)continue;
for(int j=i+1; j<=ii; j++)
{
int x1=0,x2=0;
if(j==ii)
{
for(int si=1; si<=ii; si++)
{
if(si<=i)
{
x1=x1*10+a[si];
}
else if(si<=ii)
{
x2=x2*10+a[si];
}

}
su=x1+x2;
if(su>=sum&&su<=n)
{
b[1]=x1;
b[2]=x2;
ko=2;
sum=su;

}
q[su]++;

continue;
}
int ji=i;
int si=0;
while(ji<j)
{
si=si*10+a[++ji];
}
mx=si+mi;
if(mx>n)continue;
for(int k=j+1; k<=ii; k++)
{
int x1=0,x2=0,x3=0;
if(k==ii)
{
for(int si=1; si<=ii; si++)
{
if(si<=i)
{
x1=x1*10+a[si];
}
else if(si<=j)
{
x2=x2*10+a[si];
}
else if(si<=ii)
{
x3=x3*10+a[si];
}

}
su=x1+x2+x3;
if(su>=sum&&su<=n)
{
b[1]=x1;
b[2]=x2;
b[3]=x3;
ko=3;
sum=su;

}
q[su]++;
continue;
}
int sk=0;
int ki=j;
while(ki<k)
{
sk=sk*10+a[++ki];
}
mx=sk+mi+si;
if(mx>n)continue;
for(int l=k+1; l<=ii; l++)
{
if(l==ii)
{
int x1=0,x2=0,x3=0,x4=0;
for(int si=1; si<=ii; si++)
{
if(si<=i)
{
x1=x1*10+a[si];
}
else if(si<=j)
{
x2=x2*10+a[si];
}
else if(si<=k)
{
x3=x3*10+a[si];
}
else if(si<=ii)
{
x4=x4*10+a[si];
}
}
su=x1+x2+x3+x4;
if(su>=sum&&su<=n)
{
b[1]=x1;
b[2]=x2;
b[3]=x3;
b[4]=x4;
ko=4;
sum=su;

}
q[su]++;
continue;
}
int li=0;
int lj=k;
while(lj<l)
{
li=li*10+a[++lj];
}
mx=sk+mi+si+li;
if(mx>n)continue;
for(int m=l+1; m<=ii; m++)
{
int x1=0,x2=0,x3=0,x4=0,x5=0;
if(m==ii)
{
for(int si=1; si<=ii; si++)
{
if(si<=i)
{
x1=x1*10+a[si];
}
else if(si<=j)
{
x2=x2*10+a[si];
}
else if(si<=k)
{
x3=x3*10+a[si];
}
else if(si<=l)
{
x4=x4*10+a[si];
}
else if(si<=ii)
{
x5=x5*10+a[si];
}

}
su=x1+x2+x3+x4+x5;
if(su>=sum&&su<=n)
{
b[1]=x1;
b[2]=x2;
b[3]=x3;
b[4]=x4;
b[5]=x5;
ko=5;
sum=su;

}
q[su]++;

//                                    printf("%d %d %d %d %d\n",x1,x2,x3,x4,x5);
//                                    break;
continue;
}

int mj;
int ml=l;
while(ml<m)
{
mj=mj*10+a[++ml];
}
mx=sk+mi+si+li+mj;
if(mx>sum)continue;
for(int n=m+1; n<=ii; n++)
{
int x1=0,x2=0,x3=0,x4=0,x5=0,x6=0;
for(int si=1; si<=ii; si++)
{
if(si<=i)
{
x1=x1*10+a[si];
}
else if(si<=j)
{
x2=x2*10+a[si];
}
else if(si<=k)
{
x3=x3*10+a[si];
}
else if(si<=l)
{
x4=x4*10+a[si];
}
else if(si<=m)
{
x5=x5*10+a[si];
}
else if(si<=ii)
{
x6=x6*10+a[si];
}
}
su=x1+x2+x3+x4+x5+x6;
if(su>=sum&&su<=n)
{
b[1]=x1;
b[2]=x2;
b[3]=x3;
b[4]=x4;
b[5]=x5;
b[6]=x6;
ko=6;
sum=su;

}
q[su]++;
}
}
}
}
}
}

if(q[sum]>1)printf("rejected\n");
else
{
printf("%d",sum);
// printf("ko=%d\n",ko);
// break;
for(int i=1; i<=ko; i++)
{
printf(" %d",b[i]);
}
printf("\n");
}
}
}
return 0;
}