codeforces.com/gym/102307/problem/D 分块+并查集

D. Do Not Try This Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mr. Potato Head has given you a string ss of length nn and a list of qq operations. Each operation is defined by three integers ii, aa and kk, and a character cc. To complete an operation on a string ss, you must replace sisi, si+asi+a, si+2a,…si+2a,… and si+kasi+ka with the character cc. In order to impress Mr. Potato Head, you will write a program that will perform all of this operations, in the order they are given, and print the resulting string. If this program runs in under 2 seconds, using less than 256MB of memory, then Mr. Potato Head will be in awe!

Input

The first line contains a single string ss. ss consists of lowercase English letters and has length nn (1<n≤1051<n≤105).

The second line contains a single integer qq (1≤q≤1051≤q≤105).

Each of the next qq lines contains ii, aa, kk and cc, in this order, and defines an operation. 1≤i≤n1≤i≤n, 1≤a<n1≤a<n, 0≤k<n0≤k<n and i+ka≤ni+ka≤n. cc is a lowercase English letter.

Output

Output a single line containing the resulting string after sequentially completing all the operations on the string Mr. Potato Head gave you.

Example

input

Copy

xaaaaaaaaaaaaaaaaaaa
3
4 2 8 b
6 3 4 c
10 5 2 d

output

Copy

xaabacabcdacabdbacad

 

 

分析：

从最后一个单词从后往前修改，且已经修改的不作变动，减少次数

分块，并查集和暴力

对于d 大于 根号n的，用暴力

对于d 小于 根号n的，由于d相等，所以可以将每个点的下一个点用并查集标到还没被修改过的位置

//-------------------------代码----------------------------

//#define int ll
const int N = 1e5+10;
int n,m;

struct UF {
    int fa[N];
    int find(int x) {
        if(fa[x] == x) return x;
        return fa[x] = find(fa[x]);
    }
    
    void mer(int x,int y) {
        fa[find(x)] = find(y);
    }
    
    void init(int n) {
        fo(i,1,n) fa[i] = i;
    }
}uf[350];

void solve()
{
//    cin>>n>>m;
    cin>>s;n = s.length(),s = '>' + s;
    sq = sqrt(n);
    fo(i,1,sq) uf[i].init(n+1);
    cin>>m;
    fo(j,1,m) {
        cin>>que[j].st>>que[j].d>>que[j].len>>que[j].ch;
    }
    of(j,m,1) {
        int st = que[j].st,d = que[j].d,len = que[j].len;
        char ch = que[j].ch;
        if(d > sq) {
            fo(i,0,len) {
                int id = st + i * d;
                if(!vis[id]) {
                    s[id] = ch;
                    vis[id] = 1;
                } 
            } 
        } else {
            int id = st;
            while((id - st) / d <= len && id <= n) {
                if(!vis[id]) {
                    s[id] = ch;
                    vis[id] = 1;
                } 
                ud[d].mer(id,min(n + 1,id + d));
                id = uf[d].find(id);
            }
        }
    }
    fo(i,1,n) cout<<s[i];
    cout<<<endl;
}
void main_init() {}
signed main(){
    AC();clapping();TLE;
    cout<<fixed<<setprecision(12);
    main_init();
//  while(cin>>n,n)
//  while(cin>>n>>m,n,m)
//    int t;cin>>t;while(t -- )
    solve();
//    {solve(); }
    return 0;
}

/*样例区


*/

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