A string t is called nice if a string “2017” occurs in t as a subsequence but a string “2016” doesn’t occur in t as a subsequence. For example, strings “203434107” and “9220617” are nice, while strings “20016”, “1234” and “20167” aren’t nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it’s impossible to make a string nice by removing characters, its ugliness is ?-?1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input
The first line of the input contains two integers n and q (4?≤?n?≤?200?000, 1?≤?q?≤?200?000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits ‘0’–‘9’.

The i-th of next q lines contains two integers ai and bi (1?≤?ai?≤?bi?≤?n), describing a substring in the i-th query.

Output
For each query print the ugliness of the given substring.

Examples
Input
8 3
20166766
1 8
1 7
2 8
Output
4
3
-1
Input
15 5
012016662091670
3 4
1 14
4 15
1 13
10 15
Output
-1
2
1
-1
-1
Input
4 2
1234
2 4
1 2
Output
-1
-1
Note
In the first sample:

In the first query, ugliness(“20166766”)?=?4 because all four sixes must be removed.
In the second query, ugliness(“2016676”)?=?3 because all three sixes must be removed.
In the third query, ugliness(“0166766”)?=??-?1 because it’s impossible to remove some digits to get a nice string.
In the second sample:

In the second query, ugliness(“01201666209167”)?=?2. It’s optimal to remove the first digit ‘2’ and the last digit ‘6’, what gives a string “010166620917”, which is nice.
In the third query, ugliness(“016662091670”)?=?1. It’s optimal to remove the last digit ‘6’, what gives a nice string “01666209170”.
要是以前做過這道題就好了。。南昌的題目和這道題換湯不換藥，改改順序就好了。。
如果是單次詢問的話，就直接區間dp做就好了。但是這次是多次查詢，我們就需要利用數據結構了。區間查詢，線段樹給上。
我們定義0,1,2,3,4為"",2,20,201,2017的狀態。對于每個狀態用矩陣表示：

a[i][j]代表著從狀態i轉移到狀態j所需要花費的價值。一開始把對角線上初始化為0，其余的變為inf。
假如當前位置是2的話，那么狀態轉移矩陣就變為：

從"“變為2不需要花費價值，但是從”“到”“需要花費價值為1，因為保持”"需要刪除2。0,1,7是一樣的。
但是到6的時候，保持201到201需要刪除6，花費價值為1。因為不能出現2016，所以從2017轉移也需要刪除一個價值。

因為要求最小價值，所以類似于floyed矩陣加法：
代碼如下：

#include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=2e5+100; struct node{int a[5][5];node operator+(const node &b)const//重載加法{node c;for(int i=0;i<5;i++){for(int j=0;j<5;j++){c.a[i][j]=inf;for(int k=0;k<5;k++) c.a[i][j]=min(c.a[i][j],a[i][k]+b.a[k][j]);}}return c;} }p[maxx<<2]; char s[maxx]; int n,m;inline void pushup(int cur) {p[cur]=p[cur<<1]+p[cur<<1|1]; } inline void build(int l,int r,int cur) {if(l==r){for(int i=0;i<5;i++)for(int j=0;j<5;j++) p[cur].a[i][j]=(i==j)?0:inf;if(s[l]=='2') p[cur].a[0][0]=1,p[cur].a[0][1]=0;else if(s[l]=='0') p[cur].a[1][1]=1,p[cur].a[1][2]=0;else if(s[l]=='1') p[cur].a[2][2]=1,p[cur].a[2][3]=0;else if(s[l]=='7') p[cur].a[3][3]=1,p[cur].a[3][4]=0;else if(s[l]=='6') p[cur].a[3][3]=1,p[cur].a[4][4]=1;return ;}int mid=l+r>>1;build(l,mid,cur<<1);build(mid+1,r,cur<<1|1);pushup(cur); } inline node query(int L,int R,int l,int r,int cur) {if(l<=L&&R<=r) return p[cur];int mid=L+R>>1;if(r<=mid) return query(L,mid,l,r,cur<<1);else if(l>mid) return query(mid+1,R,l,r,cur<<1|1);else return query(L,mid,l,mid,cur<<1)+query(mid+1,R,mid+1,r,cur<<1|1); } int main() {int l,r;while(~scanf("%d%d",&n,&m)){scanf("%s",s+1);build(1,n,1);while(m--){scanf("%d%d",&l,&r);node ans=query(1,n,l,r,1);if(ans.a[0][4]==inf) printf("-1\n");else printf("%d\n",ans.a[0][4]);}}return 0; }

南昌的這道題因為是9102和8102的區別，出現的位置是在第一個，我們把原來的字符串倒一下，把詢問區間換成倒置之后的區間就好了。
代碼如下：

#include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=2e5+100; struct node{int a[5][5];node operator+(const node &b)const{node c;for(int i=0;i<5;i++){for(int j=0;j<5;j++){c.a[i][j]=inf;for(int k=0;k<5;k++) c.a[i][j]=min(c.a[i][j],a[i][k]+b.a[k][j]);}}return c;} }p[maxx<<2]; char s[maxx],ss[maxx]; int n,m;inline void pushup(int cur) {p[cur]=p[cur<<1]+p[cur<<1|1]; } inline void build(int l,int r,int cur) {if(l==r){for(int i=0;i<5;i++)for(int j=0;j<5;j++) p[cur].a[i][j]=(i==j)?0:inf;if(s[l]=='2') p[cur].a[0][0]=1,p[cur].a[0][1]=0;else if(s[l]=='0') p[cur].a[1][1]=1,p[cur].a[1][2]=0;else if(s[l]=='1') p[cur].a[2][2]=1,p[cur].a[2][3]=0;else if(s[l]=='9') p[cur].a[3][3]=1,p[cur].a[3][4]=0;else if(s[l]=='8') p[cur].a[3][3]=1,p[cur].a[4][4]=1;return ;}int mid=l+r>>1;build(l,mid,cur<<1);build(mid+1,r,cur<<1|1);pushup(cur); } inline node query(int L,int R,int l,int r,int cur) {if(l<=L&&R<=r) return p[cur];int mid=L+R>>1;if(r<=mid) return query(L,mid,l,r,cur<<1);else if(l>mid) return query(mid+1,R,l,r,cur<<1|1);else return query(L,mid,l,mid,cur<<1)+query(mid+1,R,mid+1,r,cur<<1|1); } int main() {int l,r;while(~scanf("%d%d",&n,&m)){scanf("%s",ss+1);for(int i=1;i<=n;i++) s[i]=ss[n-i+1];build(1,n,1);while(m--){scanf("%d%d",&l,&r);node ans=query(1,n,n-r+1,n-l+1,1);if(ans.a[0][4]==inf) printf("-1\n");else printf("%d\n",ans.a[0][4]);}}return 0; }

努力加油a啊，(^o)/~

創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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