zoukankan      html  css  js  c++  java
  • Divide by Three CodeForces

    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
    1033
    Output
    33
    Input
    10
    Output
    0
    Input
    11
    Output
    -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.

    给定一个串,要求该数去掉尽可能少的位,使得剩下的数可以被3整除

    就是所有位的和都可以被3整除

    其实最多减两个 如果 sum%3==1 那么可以减去两个 %3==2的 和一个 %3=1的

    反之亦然

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <queue>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <set>
     7 #include <iostream>
     8 #include <map>
     9 #include <stack>
    10 #include <string>
    11 #include <vector>
    12 #define  pi acos(-1.0)
    13 #define  eps 1e-6
    14 #define  fi first
    15 #define  se second
    16 #define  lson l,m,rt<<1
    17 #define  rson m+1,r,rt<<1|1
    18 #define  bug         printf("******
    ")
    19 #define  mem(a,b)    memset(a,b,sizeof(a))
    20 #define  fuck(x)     cout<<"["<<"x="<<x<<"]"<<endl
    21 #define  f(a)        a*a
    22 #define  sf(n)       scanf("%d", &n)
    23 #define  sff(a,b)    scanf("%d %d", &a, &b)
    24 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
    25 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
    26 #define  FIN         freopen("DATA.txt","r",stdin)
    27 #define  gcd(a,b)    __gcd(a,b)
    28 #define  lowbit(x)   x&-x
    29 #pragma  comment (linker,"/STACK:102400000,102400000")
    30 using namespace std;
    31 typedef long long  LL;
    32 typedef unsigned long long ULL;
    33 const int INF = 0x7fffffff;
    34 const LL  INFLL = 0x3f3f3f3f3f3f3f3fLL;
    35 const int mod = 1e9 + 7;
    36 const int maxn = 1e4 + 10;
    37 string s1, s2, s3;
    38 int cal(string s) {
    39     int sum = 0;
    40     for (int i = 0 ; i < s.size() ; i++) sum = (sum + s[i] - '0') % 3;
    41     if (!s.size()) return 1;
    42     return sum;
    43 }
    44 void del(string &s) {
    45     while(s[0] == '0' && s.size() > 1) s.erase(0, 1);
    46 }
    47 int main() {
    48     cin >> s1;
    49     int tot = cal(s1), num1 = 1, num2 = 2, p = 3 - tot;
    50     s2 = s3 = s1;
    51     if (!tot) {
    52         cout << s1 << endl;
    53         return 0;
    54     }
    55     for (int i = s1.size() ; i >=0 ; i--) {
    56         if ((s2[i] - '0')%3 == tot && num1 ) {
    57             s2.erase(i, 1);
    58             num1--;
    59         }
    60         if ((s3[i] - '0')%3 == p  && num2) {
    61             s3.erase(i, 1);
    62             num2--;
    63         }
    64     }
    65     num1 = cal(s2), num2 = cal(s3);
    66     del(s2), del(s3);
    67     if (num1 && num2) return 0 * printf("-1
    ");
    68     if (!num1 && !num2) {
    69         if (s2.size() > s3.size()) cout << s2 << endl;
    70         else cout << s3 << endl;
    71     } else if (!num1) cout << s2 << endl;
    72     else cout << s3 << endl;
    73     return 0;
    74 }
  • 相关阅读:
    intellij idea
    this.getClass().getResource(String) 路径问题
    org.hibernate.AssertionFailure: null id 错误
    Hibernate 映射
    关于idea 在创建maven 骨架较慢问题解决
    常用base.css
    form表单样式
    ul li自适应居中导航
    table-cell实现未知宽高图片,文本水平垂直居中在div
    多行文字水平垂直居中在div
  • 原文地址:https://www.cnblogs.com/qldabiaoge/p/9527252.html
Copyright © 2011-2022 走看看