题目链接
题意
初始时有(n)个空串，然后进行(q)次操作，操作分为以下两种：
- wrap l r x：把(l,r)中的每个字符串的首尾都加入(x)，如(s_i=121,x=3 ightarrow s_i=31213)；
- query l r：查询(sumlimits_{i=l}^{r}s_i)。
思路
线段树。
对于每个结点我们存下这个区间的区间和(sum1)，区间长度(sum2)（由于区间内的数操作次数不同会导致数的位数不同所以我们用(10^x)来表示叶子结点长度，然后往上累加得到每个区间的长度），高位添加的数的懒惰标记(lazy1)，低位添加的数的懒惰标记(lazy2)，位数的懒惰标记(lazy3)，然后进行线段树区间更新(+)区间查询即可……注意理清思路别写(bug)。
代码实现如下
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
char op[10];
int t, n, q, l, r, x;
struct node {
int l, r;
int sum1, sum2;
int vis, lazy1, lazy2, lazy3;
}segtree[maxn<<2];
void push_up(int rt) {
segtree[rt].sum1 = (segtree[lson].sum1 + segtree[rson].sum1) % mod;
segtree[rt].sum2 = (segtree[lson].sum2 + segtree[rson].sum2) % mod;
}
void push_down(int rt) {
if(!segtree[rt].vis) return;
int x = segtree[rt].lazy1, y = segtree[rt].lazy2, z = segtree[rt].lazy3;
segtree[rt].vis = segtree[rt].lazy1 = segtree[rt].lazy2 = 0, segtree[rt].lazy3 = 1;
segtree[lson].vis = segtree[rson].vis = 1;
segtree[lson].lazy1 = (1LL * segtree[lson].lazy3 * x % mod + segtree[lson].lazy1) % mod;
segtree[rson].lazy1 = (1LL * segtree[rson].lazy3 * x % mod + segtree[rson].lazy1) % mod;
segtree[lson].lazy2 = (1LL * z * segtree[lson].lazy2 % mod + y) % mod;
segtree[rson].lazy2 = (1LL * z * segtree[rson].lazy2 % mod + y) % mod;
segtree[lson].sum1 = ((1LL * segtree[lson].sum2 * z % mod * x % mod + 1LL * segtree[lson].sum1 * z % mod) % mod + 1LL * y * (segtree[lson].r - segtree[lson].l + 1) % mod) % mod;
segtree[rson].sum1 = ((1LL * segtree[rson].sum2 * z % mod * x % mod + 1LL * segtree[rson].sum1 * z % mod) % mod + 1LL * y * (segtree[rson].r - segtree[rson].l + 1) % mod) % mod;
segtree[lson].sum2 = 1LL * segtree[lson].sum2 * z % mod * z % mod;
segtree[rson].sum2 = 1LL * segtree[rson].sum2 * z % mod * z % mod;
segtree[lson].lazy3 = 1LL * segtree[lson].lazy3 * z % mod;
segtree[rson].lazy3 = 1LL * segtree[rson].lazy3 * z % mod;
}
void build(int rt, int l, int r) {
segtree[rt].l = l, segtree[rt].r = r;
segtree[rt].lazy1 = segtree[rt].lazy2 = 0;
segtree[rt].lazy3 = 1;
segtree[rt].sum1 = segtree[rt].vis = 0, segtree[rt].sum2 = 1;
if(l == r) return;
int mid = (l + r) >> 1;
build(lson, l, mid);
build(rson, mid + 1, r);
push_up(rt);
}
void update(int rt, int l, int r, int x) {
if(segtree[rt].l == l && segtree[rt].r == r) {
segtree[rt].vis = 1;
segtree[rt].lazy2 = (10LL * segtree[rt].lazy2 + x) % mod;
segtree[rt].lazy1 = (1LL * segtree[rt].lazy3 * x % mod + segtree[rt].lazy1) % mod;
segtree[rt].sum1 = (10LL * ((1LL*segtree[rt].sum2 * x % mod + segtree[rt].sum1 % mod) % mod) % mod + x * (r - l + 1)) % mod;
segtree[rt].sum2 = 100LL * segtree[rt].sum2 % mod;
segtree[rt].lazy3 = 10LL * segtree[rt].lazy3 % mod;
return;
}
push_down(rt);
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
if(r <= mid) update(lson, l, r, x);
else if(l > mid) update(rson, l, r, x);
else {
update(lson, l, mid, x);
update(rson, mid + 1, r, x);
}
push_up(rt);
}
int query(int rt, int l, int r) {
if(segtree[rt].l == l && segtree[rt].r == r) return segtree[rt].sum1;
push_down(rt);
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
if(r <= mid) return query(lson, l, r);
else if(l > mid) return query(rson, l, r);
else return (query(lson, l, mid) + query(rson, mid + 1, r)) % mod;
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
scanf("%d", &t);
int icase = 0;
while(t--) {
scanf("%d%d", &n, &q);
build(1, 1, n);
printf("Case %d:
", ++icase);
while(q--) {
scanf("%s%d%d", op, &l, &r);
if(op[0] == 'w') {
scanf("%d", &x);
update(1, l, r, x);
} else printf("%d
", query(1, l, r));
}
}
return 0;
}