Educational Codeforces Round 18 C. Divide by Three DP

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原文链接：http://www.cnblogs.com/zxhl/p/6638762.html

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#数据结构与算法

本文介绍了一种使用动态规划方法来解决一个特定问题的算法：即给定一个正整数，通过删除最少数量的数字使其变为3的倍数，并确保结果没有前导零。文章详细解释了状态定义、转移方程以及实现细节。

A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.

The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.

Write a program which for the given n will find a beautiful number such that n can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number n.

If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.

Input

The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000).

Output

Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1.

Examples

input

output

Note

In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two.

题意：

　　给你一个01串，问你最少删除多少个字符，使得余下的串10进制下%3=0，不得有前导0

题解：

　　设定dp[i][j][0/1/2]表示前i个字符中，组成%3=j的串需要的最少删除次数；

　　同时0表示还未填数，

1表示有一个前导0，

　 2表示开头填了一个非0数

　　需要记录路径pre

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 1e5+10, M = 1e3+20, mod = 1e9+7, inf = 2e9;


int dp[N][5][3],pre[N][5][3];//前i个数mod3 = j最少需要删除的字母个数 是否有前导0
char s[N];
int n,a[N],ans[N];
int main() {
    scanf("%s",s+1);
    int n = strlen(s+1);
    for(int i = 1; i <= n; ++i) a[i] = s[i] - '0';
    for(int i = 0; i <= n; ++i) {
        for(int j = 0; j < 3; ++j) dp[i][j][0] = inf,dp[i][j][1] = inf, dp[i][j][2] = inf;
    }
    dp[0][0][0] = 0;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < 3; ++j) {
            if(dp[i][j][0] < dp[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2]) {
                dp[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2] = dp[i][j][0];
                pre[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2] = 1;
            }
            if(dp[i][j][1]+1 < dp[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2]) {
                dp[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2] = dp[i][j][1]+1;
                pre[i+1][(j+a[i+1])%3][(a[i+1])==0?1:2] = 1;
            }
            if(dp[i][j][2] < dp[i+1][(j+a[i+1])%3][2]) {
                dp[i+1][(j+a[i+1])%3][2] = dp[i][j][2];
                pre[i+1][(j+a[i+1])%3][2] = 1;
            }

            if(dp[i][j][1]+1 < dp[i+1][j][1]) {
               dp[i+1][j][1] = dp[i][j][1]+1;
               pre[i+1][j][1] = -1;
            }
            if(dp[i][j][0]+1 < dp[i+1][j][0]) {
               dp[i+1][j][0] = dp[i][j][0]+1;
               pre[i+1][j][0] = -1;
            }
            if(dp[i][j][2]+1 < dp[i+1][j][2]) {
               dp[i+1][j][2] = dp[i][j][2]+1;
               pre[i+1][j][2] = -1;
            }
        }
    }
    if(dp[n][0][2] >= inf && dp[n][0][1] >= inf) {
        puts("-1");
        return 0;
    }
    if(dp[n][0][1] < dp[n][0][2]) {
        puts("0");
        return 0;
    }
    int j = 0,num = n - dp[n][0][2];
    for(int i = n; i >= 1; --i) {
        if(pre[i][j][2] == 1) {
            ans[num--] = a[i];
            j = ((j - a[i])%3 + 3) % 3;
        }
        if(num == 0) break;
    }
    for(int i = 1; i <= n - dp[n][0][2]; ++i) cout<<ans[i];
    return 0;
}

转载于:https://www.cnblogs.com/zxhl/p/6638762.html