2019南昌网络赛 hello 2019

这道题和一道2017，2016的类似。

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”.
要是以前做过这道题就好了。。南昌的题目和这道题换汤不换药，改改顺序就好了。。
如果是单次询问的话，就直接区间dp做就好了。但是这次是多次查询，我们就需要利用数据结构了。区间查询，线段树给上。
我们定义0,1,2,3,4为"",2,20,201,2017的状态。对于每个状态用矩阵表示：

a[i][j]代表着从状态i转移到状态j所需要花费的价值。一开始把对角线上初始化为0，其余的变为inf。
假如当前位置是2的话，那么状态转移矩阵就变为：

从"“变为2不需要花费价值，但是从”“到”“需要花费价值为1，因为保持”"需要删除2。0,1,7是一样的。
但是到6的时候，保持201到201需要删除6，花费价值为1。因为不能出现2016，所以从2017转移也需要删除一个价值。

因为要求最小价值，所以类似于floyed矩阵加法：
————————————————
版权声明：本文为CSDN博主「starlet_kiss」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/starlet_kiss/article/details/100694910

代码如下：

 #include<bits/stdc++.h>
 #define ll long long
 #define inf 0x3f3f3f3f
 using namespace std;

 const int maxx=2e5+;
 struct node{
     int a[][];
     node operator+(const node &b)const//重载加法
     {
         node c;
         for(int i=;i<;i++)
         {
             for(int j=;j<;j++)
             {
                 c.a[i][j]=inf;
                 for(int k=;k<;k++)
                 c.a[i][j]=min(c.a[i][j],a[i][k]+b.a[k][j]);
             }
         }
         return c;
     }
 }p[maxx<<];
 char s[maxx];
 int n,m;

 inline void pushup(int cur)
 {
     p[cur]=p[cur<<]+p[cur<<|];
 }
 inline void build(int l,int r,int cur)
 {
     if(l==r)
     {
         for(int i=;i<;i++)for(int j=;j<;j++) p[cur].a[i][j]=(i==j)?:inf;
         if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         return ;
     }
     int mid=l+r>>;
     build(l,mid,cur<<);
     build(mid+,r,cur<<|);
     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>>;
     if(r<=mid) return query(L,mid,l,r,cur<<);
     else if(l>mid) return query(mid+,R,l,r,cur<<|);
     else return query(L,mid,l,mid,cur<<)+query(mid+,R,mid+,r,cur<<|);
 }
 int main()
 {
     int l,r;
     while(~scanf("%d%d",&n,&m))
     {
         scanf("%s",s+);
         build(,n,);
         while(m--)
         {
             scanf("%d%d",&l,&r);
             node ans=query(,n,l,r,);
             if(ans.a[][]==inf) printf("-1\n");
             else printf("%d\n",ans.a[][]);
         }
     }
     return ;
 }
 ————————————————
 版权声明：本文为CSDN博主「starlet_kiss」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
 原文链接：https://blog.csdn.net/starlet_kiss/article/details/100694910

南昌的这道题因为是9102和8102的区别，出现的位置是在第一个，我们把原来的字符串倒一下，把询问区间换成倒置之后的区间就好了。
代码如下：

 #include<bits/stdc++.h>
 #define ll long long
 #define inf 0x3f3f3f3f
 using namespace std;

 const int maxx=2e5+;
 struct node{
     int a[][];
     node operator+(const node &b)const
     {
         node c;
         for(int i=;i<;i++)
         {
             for(int j=;j<;j++)
             {
                 c.a[i][j]=inf;
                 for(int k=i;k<;k++){
                     if(k>j) continue;
                     c.a[i][j]=min(c.a[i][j],a[i][k]+b.a[k][j]);
                 }
             }
         }
         return c;
     }
 }p[maxx<<];
 char s[maxx],ss[maxx];
 int n,m;

 inline void pushup(int cur)
 {
     p[cur]=p[cur<<]+p[cur<<|];
 }
 inline void build(int l,int r,int cur)
 {
     if(l==r)
     {
         for(int i=;i<;i++)for(int j=;j<;j++) p[cur].a[i][j]=(i==j)?:inf;
         if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         else if(s[l]=='') p[cur].a[][]=,p[cur].a[][]=;
         return ;
     }
     int mid=l+r>>;
     build(l,mid,cur<<);
     build(mid+,r,cur<<|);
     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>>;
     if(r<=mid) return query(L,mid,l,r,cur<<);
     else if(l>mid) return query(mid+,R,l,r,cur<<|);
     else return query(L,mid,l,mid,cur<<)+query(mid+,R,mid+,r,cur<<|);
 }
 int main()
 {
     int l,r;
     while(~scanf("%d%d",&n,&m))
     {
         scanf("%s",ss+);
         for(int i=;i<=n;i++) s[i]=ss[n-i+];
         build(,n,);
         while(m--)
         {
             scanf("%d%d",&l,&r);
             node ans=query(,n,n-r+,n-l+,);
             if(ans.a[][]==inf) printf("-1\n");
             else printf("%d\n",ans.a[][]);
         }
     }
     return ;
 }