HDU 4046 Panda

线段树单点更新，要注意两段合并多出的答案的计算即可

//============================================================================
// Name        : D.cpp
// Author      : L_Ecry
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include <iostream>
#define N 50050
using namespace std;
int value[N*];
char s[N];
inline int cal(int l,int r,int mid)
{

    int res=;
    if(mid>l&&s[mid-]=='w'&&s[mid]=='b'&&s[mid+]=='w')res++;
    if(mid+<r&&s[mid]=='w'&&s[mid+]=='b'&&s[mid+]=='w')res++;
    return res;
}
void build(int l,int r,int i)
{
    if(l==r)
    {
        value[i]=;
        return ;
    }
    int mid=(l+r)>>;
    int lson=(i<<),rson=(i<<|);
    build(l,mid,lson);
    build(mid+,r,rson);
    value[i]=value[lson]+value[rson]+cal(l,r,mid);
}
void update(int l,int r,int p,int i)
{
    if(l==r)
    {
        return;
    }
    int mid=(l+r)>>;
    int lson=(i<<),rson=(i<<|);
    if(p<=mid)update(l,mid,p,lson);
    else update(mid+,r,p,rson);
    value[i]=value[lson]+value[rson]+cal(l,r,mid);
}
int query(int l,int r,int pl,int pr,int i)
{
    if(l==pl&&r==pr)
    {
        return value[i];
    }
    int mid=(l+r)>>;
    if(pr<=mid)return query(l,mid,pl,pr,i<<);
    else if(pl>mid)return query(mid+,r,pl,pr,i<<|);
    else
    {
        return query(l,mid,pl,mid,i<<)+query(mid+,r,mid+,pr,i<<|)+cal(pl,pr,mid);
    }
}
int n,m;
void init()
{
    scanf("%d%d",&n,&m);
    scanf(" %s",s+);
    build(,n,);
}
void solve()
{
    while(m--)
    {
        int x,y,z;
        char c;
        scanf("%d%d",&x,&y);
        if(x==)
        {
            scanf("%d",&z);
            y++;z++;
            printf("%d\n",query(,n,y,z,));
        }
        else
        {
            scanf(" %c",&c);
            y++;
            s[y]=c;
            update(,n,y,);
        }
    }
}
int main() {
    int tt,ri=;
    scanf("%d",&tt);
    while(tt--)
    {
        init();
        printf("Case %d:\n",++ri);
        solve();
    }
    return ;
}