有奖求调！！！

T3 求调，12：00 之前调成功悬一袋红烧牛肉面

#include<bits/stdc++.h>
#define Type int
#define qr(x) x=read()
typedef long long ll;
using namespace std;inline Type read(){char c=getchar(); Type x=0;while(!isdigit(c))c=getchar();while(isdigit(c))x=(x<<1)+(x<<3)+(c^48),c=getchar();return x;
}const int N = 2e5 + 10;
const int mod = 1e9 + 7;int n, q, f[N];
char s[N];int P(char A, char B){if(A == '#' or B == '@') return 1;if(A == B) return 0;if(A == 'R'){if(B == 'S') return 1;else return -1;}else if(A == 'S'){if(B == 'P') return 1;else return -1;}else{if(B == 'R') return 1;else return -1;}
}
struct tree{int v, pos, tag;bool operator < (const tree &A) const{return v < A.v;}}t[N<<3];
namespace Tree
{#define lson rt<<1#define rson rt<<1|1inline void pushup(int rt){if(t[lson].v < t[rson].v) t[rt].v = t[lson].v, t[rt].pos = t[lson].pos;else t[rt].v = t[rson].v, t[rt].pos = t[rson].pos;}inline void pushdown(int rt){if(t[rt].tag){t[lson].tag += t[rt].tag, t[rson].tag += t[rt].tag;t[lson].v += t[rt].tag, t[rson].v += t[rt].tag;t[rt].tag = 0;}}inline void build(int rt, int l, int r){if(l == r){t[rt].v = f[l], t[rt].pos = l;return;}int mid = (l + r) >> 1;build(lson, l, mid),build(rson, mid+1, r);pushup(rt);// if(t[rt].pos == 0) cout<<"CTHisSB\n";}inline void update(int rt, int l, int r, int pos, int val){if(pos <= l and r <= n){t[rt].tag += val, t[rt].v += val;return;}pushdown(rt);int mid = (l + r) >> 1;if(pos <= mid) update(lson, l, mid, pos, val);if(n > mid) update(rson, mid+1, r, pos, val);pushup(rt);}inline tree query(int rt, int l, int r, int L, int R){if(L <= l and r <= R) return t[rt];pushdown(rt); int mid = (l + r) >> 1;if(R <= mid) return query(lson, l, mid, L, R);else{if(L > mid) return query(rson, mid+1, r, L, R);else return min(query(lson, l, mid, L, R), query(rson, mid+1, r, L, R));}}
}signed main(){freopen("a.in", "r", stdin), freopen("a.out", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);qr(n), qr(q), cin>>(s+1);f[1] = 1; s[0] = '#', s[n+1] = '@';for(int i=2; i<=n; i++)f[i] = f[i-1] + P(s[i-1], s[i]);Tree::build(1, 1, n);while(q--){int qr(op);switch(op){case 1:{int qr(p); char c; cin>>c; int now = P(s[p-1], c) - P(s[p-1], s[p]);Tree::update(1, 1, n, p, now); now = P(c, s[p+1]) - P(s[p], s[p+1]);Tree::update(1, 1, n, p+1, now); s[p] = c;  break; }default:{int qr(l), qr(r); cout<<s[Tree::query(1, 1, n, l, r).pos]<<"\n"; break;}}}return 0;
}

为了方便判断，我让 s[0]='#'，但交到 oj 上却输出了这个：

显然是返回了 0，但是毫无缘由，代码里注释部分是输出树上所有点的 pos，结果没有为 0 的。