Codeforces Round #545 (Div. 2) D. Camp Schedule

https://codeforces.com/contest/1138/problem/D

看了半天,才发现要用kmp,不过kmp一个多月前学会了就没用过,都不会写了,又复习了一遍。

思路:利用kmp的next数组,可以知道按最长前缀后缀一样的长度来输出,最后再补上剩下的,没什么难的,就是南想到kmp

#include 
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define DEBUG cout< pii;
const int maxn = 3e5 + 200;
string s, b;
int cnt0, cnt1, net[1000010];
void getnetval(string p) {
    int i = 1, j = 0;
    net[0] = -1;
    while(i < p.size()) {
        if(j == -1 || p[i] == p[j]) {
            ++i, ++j;
            net[i] = j;
        } else
            j = net[j];
    }
}
int KmpSearch(char* s, char* p) {
    int i = 0, j = 0;
    int sLen = strlen(s);
    int pLen = strlen(p);
    while (i < sLen && j < pLen) {
        if (j == -1 || s[i] == p[j])
            i++, j++;
        else
            j = net[j];
    }
    return j == pLen ? i - j : -1;
}
int main() {
    cin >> s >> b;
    getnetval(b);
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == '0')
            cnt0++;
        else
            cnt1++;
    }
    int l = net[b.size()];
    int flag = 0;
    for (int i = 0; i < l; i++) {
        if (b[i] == '0' && cnt0 > 0) {
            cout << 0;
            cnt0--;
        } else if (b[i] == '1' && cnt1 > 0) {
            cout << 1;
            cnt1--;
        } else {
            flag = 1;
            break;
        }
    }
    if (!flag)
        while (cnt0 > 0 || cnt1 > 0) {
            flag = 0;
            for (int i = l; i < b.size(); i++) {
                if (b[i] == '0' && cnt0 > 0) {
                    cout << 0;
                    cnt0--;
                } else if (b[i] == '1' && cnt1 > 0) {
                    cout << 1;
                    cnt1--;
                } else {
                    flag = 1;
                    break;
                }
            }
            if (flag)
                break;
        }
    for (int i = 0; i < cnt0; i++)
        cout << 0;
    for (int i = 0; i < cnt1; i++)
        cout << 1;
    return 0;
}

 

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