HDU4399：Query

Problem Description

You are given two strings s1[0..l1], s2[0..l2] and Q - number of queries.
Your task is to answer next queries:
  1) 1 a i c - you should set i-th character in a-th string to c;
  2) 2 i - you should output the greatest j such that for all k (i<=k and k<i+j) s1[k] equals s2[k].

 

Input

The first line contains T - number of test cases (T<=25).
Next T blocks contain each test.
The first line of test contains s1.
The second line of test contains s2.
The third line of test contains Q.
Next Q lines of test contain each query:
  1) 1 a i c (a is 1 or 2, 0<=i, i<length of a-th string, 'a'<=c, c<='z')
  2) 2 i (0<=i, i<l1, i<l2)
All characters in strings are from 'a'..'z' (lowercase latin letters).
Q <= 100000.
l1, l2 <= 1000000.

 

Output

For each test output "Case t:" in a single line, where t is number of test (numbered from 1 to T).
Then for each query "2 i" output in single line one integer j.

 

Sample Input

1aaabbaaabbaa72 02 12 22 31 1 2 b2 02 3

 

Sample Output

Case 1:210141

 

//苦逼

看了很久才看懂

果然我还是太水了

先说 1 1 2 b

意思就是把 1 字符串的下标为 2 的字符改为 b

这样1的字符串就变为饿aabbba

然后2 0

就是两个字符串从0下标开始

算下连续有几个相同的字符

差不多就这样了

#include <iostream>#include <algorithm>#include <cstdio>#include <memory.h>using namespace std;const int maxn = 1000007;int flag[maxn<<2],sum[maxn<<2],t,cas = 1;char s1[maxn],s2[maxn];#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1void pushUP(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void build(int l,int r,int rt){    if(l == r)    {        if(s1[l] == s2[l]) flag[l] = 1;        else flag[l] = 0;        sum[rt] = flag[l];        return ;    }    int m = (l+r)>>1;    build(lson);    build(rson);    pushUP(rt);}void update(int op,int pos,char c,int l,int r,int rt){    if(l == r)    {        if(op == 1) s1[l] = c;        if(op == 2) s2[l] = c;        if(s1[l] == s2[l]) flag [l] = 1;        else flag[l] = 0;        sum[rt] = flag[l];        return ;    }    int m = (l+r)>>1;    if(pos<=m) update(op,pos,c,lson);    else update(op,pos,c,rson);    pushUP(rt);}int query(int pos,int l,int r,int rt){    if(l == r) return sum[rt];    if(l<=pos && pos<=r)    {        if(sum[rt] == r-l+1)        {            return r-pos+1;        }        else        {            int m = (l+r)>>1,ans = 0;            if(pos <= m)            {                ans+=query(pos,lson);                if(ans == m-pos+1)                {                    ans+=query(m+1,rson);                }            }            else ans+=query(pos,rson);            return ans;        }    }}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%s%s",(s1+1),(s2+1));        int l1 = strlen(s1+1);        int l2 = strlen(s2+1);        int n = max(l1,l2);        build(1,n,1);        int m,op,idx,pos;        char c[2];        scanf("%d",&m);        printf("Case %d:\n",cas++);        while(m--)        {            scanf("%d",&op);            if(op == 2)            {                scanf("%d",&pos);                printf("%d\n",query(pos+1,1,n,1));            }            else            {                scanf("%d%d%s",&idx,&pos,&c);                update(idx,pos+1,c[0],1,n,1);            }        }    }    return 0;}