大意: 给定01串, 单点修改, 询问给定区间$[l,r]$, 假设$[l,r]$从左往右得到的二进制数为$x$, 每次操作增加或减少2的幂, 求最少操作数使得$x$为0.
线段树维护2*2矩阵表示低位是否进位,高位是否进位的最少花费.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n, t; char s[N]; struct _ { int v[2][2]; void upd(int x) { if (x==1) v[0][0]=v[0][1]=v[1][0]=1,v[1][1]=0; else v[0][0]=0,v[0][1]=v[1][0]=v[1][1]=1; } _ operator + (const _ &rhs) const { _ ret; memset(ret.v,0x3f,sizeof ret.v); REP(k,0,1) REP(i,0,1) REP(j,0,1) { ret.v[i][j]=min(ret.v[i][j],v[i][k]+rhs.v[k][j]); } return ret; } } tr[N<<2]; void pu(int o) { tr[o]=tr[lc]+tr[rc]; } void build(int o, int l, int r) { l==r?tr[o].upd(s[l]-'0'):(build(ls),build(rs),pu(o)); } void update(int o, int l, int r, int x, int v) { l==r?tr[o].upd(v):(mid>=x?update(ls,x,v):update(rs,x,v),pu(o)); } _ qry(int o, int l, int r, int ql, int qr) { if (ql<=l&&r<=qr) return tr[o]; if (mid>=qr) return qry(ls,ql,qr); if (mid<ql) return qry(rs,ql,qr); return qry(ls,ql,qr)+qry(rs,ql,qr); } int main() { scanf("%d%s", &n, s+1); build(1,1,n); scanf("%d", &t); while (t--) { int op, x, y; scanf("%d%d%d", &op, &x, &y); if (op==1) printf("%d ",qry(1,1,n,x,y).v[0][0]); else update(1,1,n,x,y); } }