++ 上一個 sub stream out cti then head rac

Polycarp and Div 3 time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

Polycarp likes numbers that are divisible by 3.

He has a huge number $\mathrm{ss.\; Polycarp\; wants\; to\; cut\; from\; it\; the\; maximum\; number\; of\; numbers\; that\; are\; divisible\; by\; 33.\; To\; do\; this,\; he\; makes\; an\; arbitrary\; number\; of\; vertical\; cuts\; between\; pairs\; of\; adjacent\; digits.\; As\; a\; result,\; after}$

mm such cuts, there will be $\mathrm{m+1m+1\; parts\; in\; total.\; Polycarp\; analyzes\; each\; of\; the\; obtained\; numbers\; and\; finds\; the\; number\; of\; those\; that\; are\; divisible\; by\; 33.}$

For example, if the original number is $\mathrm{s=3121s=3121,\; then\; Polycarp\; can\; cut\; it\; into\; three\; parts\; with\; two\; cuts:\; 3|1|213|1|21.\; As\; a\; result,\; he\; will\; get\; two\; numbers\; that\; are\; divisible\; by}$

33.

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character ‘0‘). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by $\mathrm{33\; that\; Polycarp\; can\; obtain?}$

Input

The first line of the input contains a positive integer $\mathrm{ss.\; The\; number\; of\; digits\; of\; the\; number\; ss\; is\; between\; 11\; and\; 2?1052?105,\; inclusive.\; The\; first\; (leftmost)\; digit\; is\; not\; equal\; to\; 0.}$

Output

Print the maximum number of numbers divisible by $\mathrm{33\; that\; Polycarp\; can\; get\; by\; making\; vertical\; cuts\; in\; the\; given\; number\; ss.}$

Examples input Copy

3121

output Copy

2

input Copy

6

output Copy

1

input Copy

1000000000000000000000000000000000

output Copy

33

input Copy

201920181

output Copy

4

Note

In the first example, an example set of optimal cuts on the number is 3|1|21.

In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $33.$

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $\mathrm{3333\; digits\; 0.\; Each\; of\; the\; 3333\; digits\; 0\; forms\; a\; number\; that\; is\; divisible\; by\; 33.}$

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $\mathrm{00,\; 99,\; 201201\; and\; 8181\; are\; divisible\; by\; 33.}$

題意：給我們一個數，我們可以對這個數進行任意的劃分，但是不能出現前綴零的無意義數，問我們最多可以劃分出幾個可以整除三的數？

分析：直接遍歷，考慮這四種情況就可以

1.如果單獨的數能整除三，那麽這數肯定可以，結果直接加一

2.如果不能整除三，那麽將得到一個余數，然後接下來我們會在這個數的基礎上加上新的數，如果余數和這個數的余數相同，則中間肯定加了一個能整除三的數，結果加一

3.如果不能整除三的數加上一個數後能整除三，則結果加一

4.如果不能整除三的數加上一個數既不能整除三且余數和之前不相同，則再加一個數肯定會得到和前面兩個數相同的余數或者整除三

根據以上分析我們每次加數時將余數打上標誌，然後根據上訴分析判斷結果是否能加一，記得結果加一後要把標誌置為0

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int main() {
    string s;
    while( cin >> s ) {
        ll now = 0, cnt = 0, vis[3] = { 1, 0, 0 };
        for( ll i = 0; i < s.length(); i ++ ) {
            now += s[i]-‘0‘;
            now %= 3;
            if( vis[now] ) {
                cnt ++;
                now = 0;
                vis[1] = vis[2] = 0;
            } else {
                vis[now] = 1;
            }
        }
        cout << cnt << endl;
    }
    return 0;
}

CF1005D Polycarp and Div 3 思維