Discription

This task is very simple. Given a string?S?of length?n?and?q?queries each query is on the format?i?j?k?which means sort the substring consisting of the characters from?i?to?j?in non-decreasing order if?k?=?1?or in non-increasing order if?k?=?0.

Output the final string after applying the queries.

Input

The first line will contain two integers?n,?q?(1?≤?n?≤?10⁵,?0?≤?q?≤?50?000), the length of the string and the number of queries respectively.

Next line contains a string?S?itself. It contains only lowercase English letters.

Next?q?lines will contain three integers each?i,?j,?k?(1?≤?i?≤?j?≤?n,?).

Output

Output one line, the string?S?after applying the queries.

Example

Input 10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1 Output cbcaaaabdd Input 10 1
agjucbvdfk
1 10 1 Output abcdfgjkuv

Note

First sample test explanation:

?

?

大概是第一次坐到沙發hhh

?

可以發現只有26個小寫英文，那么就不難了，就是一個常數*26的帶標記線段樹。。。

#include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<cmath> #include<algorithm> #define ll long long #define maxn 100005 using namespace std; struct node{int sum[28];int tag;node operator +(const node& u)const{node r; r.tag=0;for(int i=1;i<27;i++) r.sum[i]=sum[i]+u.sum[i];return r;} }a[maxn<<2|1]; int le,ri,opt,w; int n,m,v,num[30]; char s[maxn];void build(int o,int l,int r){if(l==r){a[o].sum[s[l]-'a'+1]++;return;}int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;build(lc,l,mid),build(rc,mid+1,r);a[o]=a[lc]+a[rc]; }inline void tol(int o,int tmp,int len){memset(a[o].sum,0,sizeof(a[o].sum));a[o].sum[tmp]=len;a[o].tag=tmp; }inline void pushdown(int o,int l,int r){if(a[o].tag){int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;tol(lc,a[o].tag,mid-l+1);tol(rc,a[o].tag,r-mid);a[o].tag=0;} }void update(int o,int l,int r){if(l>=le&&r<=ri){tol(o,v,r-l+1);return;}pushdown(o,l,r);int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;if(le<=mid) update(lc,l,mid);if(ri>mid) update(rc,mid+1,r);a[o]=a[lc]+a[rc]; }int query(int o,int l,int r){if(l>=le&&r<=ri) return a[o].sum[v];pushdown(o,l,r);int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1,an=0;if(le<=mid) an+=query(lc,l,mid);if(ri>mid) an+=query(rc,mid+1,r);return an; }void print(int o,int l,int r){if(l==r){for(int i=1;i<27;i++) if(a[o].sum[i]){putchar(i-1+'a');break;}return;}pushdown(o,l,r);int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;print(lc,l,mid);print(rc,mid+1,r); }int main(){scanf("%d%d",&n,&m);scanf("%s",s+1);build(1,1,n); while(m--){scanf("%d%d%d",&le,&ri,&opt);memset(num,0,sizeof(num));for(v=1;v<=26;v++) num[v]=query(1,1,n);if(opt){int pre=le,nxt;for(int i=1;i<=26;i++) if(num[i]){nxt=pre+num[i]-1,v=i;le=pre,ri=nxt;update(1,1,n);pre=nxt+1;}}else{int pre=le,nxt;for(int i=26;i;i--) if(num[i]){nxt=pre+num[i]-1,v=i;le=pre,ri=nxt;update(1,1,n);pre=nxt+1;} } // print(1,1,n);}print(1,1,n);return 0; }

　　

轉載于:https://www.cnblogs.com/JYYHH/p/8407093.html

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