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  • [codeforces792C][dp]

    https://codeforc.es/contest/792/problem/C

    C. Divide by Three
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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 beautifulnumber 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
    Copy
    1033
    output
    Copy
    33
    input
    Copy
    10
    output
    Copy
    0
    input
    Copy
    11
    output
    Copy
    -1
    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.

    给一个数字，你可以删除字符串某一个位置的字符，使其满足下列条件：

    • 数字没有前导0
    • 数字能够被3整除

    求经过最少操作次数之后得到的结果。

    题解：dp过程记录操作即可

      1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define debug(x) cout<<"["<<#x<<"]"<<" is "<<x<<endl;
  4 typedef long long ll;
  5 int dp[100005][4][2],pre[100005][4][2][2],xx[100005][4][2];
  6 char ch[100005],q[100005];
  7 int main(){
  8     scanf("%s",ch+1);
  9     int len=strlen(ch+1);
 10     memset(dp,-1,sizeof(dp));
 11     dp[0][0][0]=0;
 12     int f=-1;
 13     for(int i=1;i<=len;i++){
 14         int x=ch[i]-'0';
 15         if(x){
 16             if((dp[i-1][1][1]!=-1)&&(dp[i][1][1]==-1||(dp[i-1][1][1]+1<dp[i][1][1]))){
 17                 dp[i][1][1]=dp[i-1][1][1]+1;
 18                 pre[i][1][1][0]=1;
 19                 pre[i][1][1][1]=1;
 20                 xx[i][1][1]=1;
 21             }
 22             if((dp[i-1][(1-x+30)%3][1]!=-1)&&(dp[i][1][1]==-1||(dp[i-1][(1-x+30)%3][1]<dp[i][1][1]))){
 23                 dp[i][1][1]=dp[i-1][(1-x+30)%3][1];
 24                 pre[i][1][1][0]=(1-x+30)%3;
 25                 pre[i][1][1][1]=1;
 26                 xx[i][1][1]=0;
 27             }
 28             if((dp[i-1][(1-x+30)%3][0]!=-1)&&(dp[i][1][1]==-1||(dp[i-1][(1-x+30)%3][0]<dp[i][1][1]))){
 29                 dp[i][1][1]=dp[i-1][(1-x+30)%3][0];
 30                 pre[i][1][1][0]=(1-x+30)%3;
 31                 pre[i][1][1][1]=0;
 32                 xx[i][1][1]=0;
 33             }
 34             if((dp[i-1][2][1]!=-1)&&(dp[i][2][1]==-1||(dp[i-1][2][1]+1<dp[i][2][1]))){
 35                 dp[i][2][1]=dp[i-1][2][1]+1;
 36                 pre[i][2][1][0]=2;
 37                 pre[i][2][1][1]=1;
 38                 xx[i][2][1]=1;
 39             }
 40             if((dp[i-1][(2-x+30)%3][1]!=-1)&&(dp[i][2][1]==-1||(dp[i-1][(2-x+30)%3][1]<dp[i][2][1]))){
 41                 dp[i][2][1]=dp[i-1][(2-x+30)%3][1];
 42                 pre[i][2][1][0]=(2-x+30)%3;
 43                 pre[i][2][1][1]=1;
 44                 xx[i][2][1]=0;
 45             }
 46             if((dp[i-1][(2-x+30)%3][0]!=-1)&&(dp[i][2][1]==-1||(dp[i-1][(2-x+30)%3][0]<dp[i][2][1]))){
 47                 dp[i][2][1]=dp[i-1][(2-x+30)%3][0];
 48                 pre[i][2][1][0]=(2-x+30)%3;
 49                 pre[i][2][1][1]=0;
 50                 xx[i][2][1]=0;
 51             }
 52             if((dp[i-1][0][0]!=-1)&&(dp[i][0][0]==-1||(dp[i-1][0][0]+1<dp[i][0][0]))){
 53                 dp[i][0][0]=dp[i-1][0][0]+1;
 54                 pre[i][0][0][0]=0;
 55                 pre[i][0][0][1]=0;
 56                 xx[i][0][0]=1;
 57             }
 58             if((dp[i-1][0][1]!=-1)&&(dp[i][0][1]==-1||(dp[i-1][0][1]+1<dp[i][0][1]))){
 59                 dp[i][0][1]=dp[i-1][0][1]+1;
 60                 pre[i][0][1][0]=0;
 61                 pre[i][0][1][1]=1;
 62                 xx[i][0][1]=1;
 63             }
 64             if((dp[i-1][(0-x+30)%3][1]!=-1)&&(dp[i][0][1]==-1||(dp[i-1][(0-x+30)%3][1]<dp[i][0][1]))){
 65                 dp[i][0][1]=dp[i-1][(0-x+30)%3][1];
 66                 pre[i][0][1][0]=(0-x+30)%3;
 67                 pre[i][0][1][1]=1;
 68                 xx[i][0][1]=0;
 69             }
 70             if((dp[i-1][(0-x+30)%3][0]!=-1)&&(dp[i][0][1]==-1||(dp[i-1][(0-x+30)%3][0]<dp[i][0][1]))){
 71                 dp[i][0][1]=dp[i-1][(0-x+30)%3][0];
 72                 pre[i][0][1][0]=(0-x+30)%3;
 73                 pre[i][0][1][1]=0;
 74                 xx[i][0][1]=0;
 75             }
 76         }
 77         else{
 78             f=0;
 79             if((dp[i-1][0][0]!=-1)&&(dp[i][0][0]==-1||(dp[i-1][0][0]+1<dp[i][0][0]))){
 80                 dp[i][0][0]=dp[i-1][0][0]+1;
 81                 pre[i][0][0][0]=0;
 82                 pre[i][0][0][1]=0;
 83                 xx[i][0][0]=1;
 84             }
 85             if((dp[i-1][0][1]!=-1)&&(dp[i][0][1]==-1||(dp[i-1][0][1]<dp[i][0][1]))){
 86                 dp[i][0][1]=dp[i-1][0][1];
 87                 pre[i][0][1][0]=0;
 88                 pre[i][0][1][1]=1;
 89                 xx[i][0][1]=0;
 90             }
 91             if((dp[i-1][1][1]!=-1)&&(dp[i][1][1]==-1||(dp[i-1][1][1]<dp[i][1][1]))){
 92                 dp[i][1][1]=dp[i-1][1][1];
 93                 pre[i][1][1][0]=1;
 94                 pre[i][1][1][1]=1;
 95                 xx[i][1][1]=0;
 96             }
 97             if((dp[i-1][2][1]!=-1)&&(dp[i][2][1]==-1||(dp[i-1][2][1]<dp[i][2][1]))){
 98                 dp[i][2][1]=dp[i-1][2][1];
 99                 pre[i][2][1][0]=2;
100                 pre[i][2][1][1]=1;
101                 xx[i][2][1]=0;
102             }
103         }
104     }
105         int tot=0;
106         if(dp[len][0][1]!=-1&&dp[len][0][1]<dp[len][0][0]){
107         int t1=0;
108         int t2=1;
109         for(int i=len;i>=1;i--){
110             if(!xx[i][t1][t2]){
111                 q[++tot]=ch[i];
112             }
113             int tt1=t1;
114             int tt2=t2;
115             t1=pre[i][tt1][tt2][0];
116             t2=pre[i][tt1][tt2][1];
117         }
118         for(int i=tot;i>=1;i--){
119             printf("%c",q[i]);
120         }
121         printf("
");
122     }
123     else{
124         printf("%d
",f);
125     }
126     return 0;
127 }
    View Code
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  • 原文地址：https://www.cnblogs.com/MekakuCityActor/p/10801612.html
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