hdu 6562 Lovers (线段树)

大意: 有$n$个数字串, 初始为空, 两种操作(1)把$[l,r]$范围的所有数字串首位添加数位$d$ (2)询问$[l,r]$区间和

假设添加的数为$L$, $L$位数为$H$, $L$翻转后乘上$10^H$为$R$

假设$x$的位数为$h$, 那么$x$就会变为$R10^h+x10^H+L$

区间和$s_1$变为$R\sum 10^h+s_110^H+L(r-l+1)$.

再维护一个$s_2=\sum 10^h$即可

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#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 '\n'
#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, 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 = 1e5+10;
struct _ {
    int s1,s2,pre,suf,h;
    void upd(int H, int L, int R, int len) {
        s1 = ((ll)s1*H+(ll)len*L+(ll)s2*R)%P;
        s2 = (ll)s2*H%P*H%P;
        pre = ((ll)pre*H+L)%P;
        suf = ((ll)suf*H+(ll)R*h%P*h)%P;
        h = (ll)h*H%P;
    }
    _ operator + (const _ &rhs) const {
        _ ret;
        ret.s1 = (s1+rhs.s1)%P;
        ret.s2 = (s2+rhs.s2)%P;
        ret.pre = ret.suf = 0;
        ret.h = 1;
        return ret;
    }
} tr[N<<2];
void pd(int o, int l, int r) {
    if (tr[o].h!=1) {
        tr[lc].upd(tr[o].h,tr[o].pre,tr[o].suf,mid-l+1);
        tr[rc].upd(tr[o].h,tr[o].pre,tr[o].suf,r-mid);
        tr[o].h=1,tr[o].pre=tr[o].suf=0;
    }
}

void add(int o, int l, int r, int ql, int qr, int d) {
    if (ql<=l&&r<=qr) return tr[o].upd(10,d,d*10,r-l+1);
    pd(o,l,r);
    if (mid>=ql) add(ls,ql,qr,d);
    if (mid<qr) add(rs,ql,qr,d);
    tr[o] = tr[lc]+tr[rc];
}
int query(int o, int l, int r, int ql, int qr) {
    if (ql<=l&&r<=qr) return tr[o].s1;
    pd(o,l,r);
    int ans = 0;
    if (mid>=ql) ans+=query(ls,ql,qr);
    if (mid<qr) ans+=query(rs,ql,qr);
    return ans%P;
}

void build(int o, int l, int r) {
    if (l==r) tr[o] = {0,1,0,0,1};
    else build(ls),build(rs),tr[o]=tr[lc]+tr[rc];
}
int n, m;

void work() {
    scanf("%d%d",&n,&m);
    build(1,1,n);
    while (m--) {
        char s[10];
        int l,d,r;
        scanf("%s%d%d",s,&l,&r);
        if (s[0]=='w') scanf("%d",&d),add(1,1,n,l,r,d);
        else printf("%d\n",query(1,1,n,l,r));
    }
}

int main() {
    int t=rd();
    REP(i,1,t) {
        printf("Case %d:\n",i);
        work();
    }
}