uav 11258 String Partition (DP)

Problem F - String Partition
                                                                                                                   
Time limit: 3.000 seconds



John was absurdly busy for preparing a programming contest recently. He wanted to create a ridiculously easy problem for the contest. His problem was not only easy, but also boring: Given a list of non-negative integers, what is the sum of them?

However, he made a very typical mistake when he wrote a program to generate the input data for his problem. He forgot to print out spaces to separate the list of integers. John quickly realized his mistake after looking at the generated input
file because each line is simply a string of digits instead of a list of integers.



He then got a better idea to make his problem a little more interesting: There are many ways to split a string of digits into a list of non-zero-leading (0 itself is allowed) 32-bit signed integers.
What is the maximum sum of the resultant integers if the string is split appropriately?



Input

The input begins with an integer N (≤ 500) which indicates the number of test cases followed. Each of the following test
cases consists of a string of at most 200 digits.



Output

For each input, print out required answer in a single line.



Sample input

6

1234554321

5432112345

000

121212121212

2147483648

11111111111111111111111111111111111111111111111111111

Sample output

1234554321

543211239

0

2121212124

214748372

5555555666

 

 

题意：给定一个字符串序列。让你把它划分成几个int型的数字，而且使这些数字之和最大。

 

思路：dp[i]表示前i个字符划分时的和的最大值。

dp[i] = max(dp[i], dp[i-j]+num)，num为从后往前划分为j长度的数字。

 

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <climits>
#define LL long long
using namespace std;
const LL inf=INT_MAX;
const int maxn=210;

LL dp[maxn];
string str;

void initial()
{
    memset(dp,0,sizeof(dp));
}

void input()
{
    cin>>str;
}

int get_num(char ch)
{
    return ch-'0';
}

void solve()
{
    int len=str.size();
    dp[1]=get_num(str[0]);

    for(int i=2;i<=len;i++)
    {
         LL tmp=1,now=0;
         for(int k=i-1,j=1;k>=0;k--,j++)
         {
              now=get_num(str[k])*tmp+now;
              if(now>inf)  break;
              dp[i]=max(dp[i],dp[i-j]+now);
              tmp*=10;
         }
    }
    printf("%lld\n",dp[len]);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
         initial();
         input();
         solve();
    }
    return 0;
}