2013
12-12

Pairs of integers

You are to find all pairs of integers such that their sum is equal to the given integer number N and the second number results from the first one by striking out one of its digits. The first integer always has at least two digits and starts with a non-zero digit. The second integer always has one digit less than the first integer and may start with a zero digit.

The input file consists of several test cases.

Each test case contains single integer N (10 ≤ N ≤ 10^9), as decribed above

The output consists of several blocks, one for each test case.

On the first line of a block write the total number of different pairs of integers that satisfy the problem statement. On the following lines write all those pairs. Write one pair on a line in ascending order of the first integer in the pair. Each pair must be written in the following format:

X + Y = N

Here X, Y, and N, must be replaced with the corresponding integer numbers. There should be exactly one space on both sides of ‘+’ and ‘=’ characters.

302

5
251 + 51 = 302
275 + 27 = 302
276 + 26 = 302
281 + 21 = 302
301 + 01 = 302

#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 5005
using namespace std;
int f[MAXN][3],n,ans;
char s[MAXN];
void getf()
{
n=strlen(s),ans=0;
for(int k=0;k<n;++k)
for(int i=0;i+k<n;++i)
{
int j=i+k,now=k%3,pre2=(now-2+3)%3;//pre记录的是f(i,j)的上两层，这里的f[i+1][pre2]指的是f(i+1,j-1)
f[i][now]=(k==0 || (k==1 && s[i]==s[j]) || (i+1<=j-1 && s[i]==s[j] && f[i+1][pre2]));
//k==0还有k=1都要特殊处理，防止i+1>j-1

ans+=f[i][now];
}
}
int main()
{
while(~scanf("%s",s))
{
getf();
cout<<ans<<endl;
}
}

1. 有一点问题。。后面动态规划的程序中
int dp[n+1][W+1];
会报错 提示表达式必须含有常量值。该怎么修改呢。。