hdu 5737(线段树)

#include <bits/stdc++.h>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;#define LL long long#define pii pair<int,int>#define MP make_pair#define ls i << 1#define rs ls | 1#define md (ll + rr >> 1)#define lson ll, md, ls#define rson md + 1, rr, rs#define Pi acos(-1.0)#define mod 1000000007#define eps 1e-12#define inf 0x3f3f3f3f#define N 100010int down[N<<2], san[N*20], A[N], B[N], sum[N<<2];int st[N<<2], ed[N<<2], to[2][N*20];int tot, n, m, a, b;void build(int ll, int rr, int i){    down[i] = -1;    san[tot++] = -inf;    st[i] = tot;    for(int j = ll; j <= rr; ++j){        san[tot++] = B[j];    }    ed[i] = tot;    sort(san + st[i], san + ed[i]);    if(ll == rr){        if(A[ll] >= B[ll]) sum[i] = 1;        else sum[i] = 0;        return ;    }    build(lson);    build(rson);    int lx = st[ls], rx = st[rs];    for(int j = st[i]; j < ed[i]; ++j){        while(lx < ed[ls] && san[lx] <= san[j]) ++lx;        while(rx < ed[rs] && san[rx] <= san[j]) ++rx;        to[0][j] = lx - 1;        to[1][j] = rx - 1;    }    to[0][st[i]-1] = st[ls] - 1;    to[1][st[i]-1] = st[rs] - 1;    sum[i] = sum[ls] + sum[rs];}void push_down(int i){    if(~down[i]){        down[ls] = to[0][down[i]];        sum[ls] = down[ls] - st[ls] + 1;        down[rs] = to[1][down[i]];        sum[rs] = down[rs] - st[rs] + 1;        down[i] = -1;    }}void update(int l, int r, int x, int ll, int rr, int i){    if(l == ll && r == rr){        sum[i] = x - st[i] + 1;        down[i] = x;        return ;    }    push_down(i);    if(r <= md) update(l, r, to[0][x], lson);    else if(l > md) update(l, r, to[1][x], rson);    else update(l, md, to[0][x], lson), update(md + 1, r, to[1][x], rson);    sum[i] = sum[ls] + sum[rs];}int query(int l, int r, int ll, int rr, int i){    if(l == ll && r == rr) return sum[i];    push_down(i);    if(r <= md) return query(l, r, lson);    else if(l > md) return query(l, r, rson);    else return query(l, md, lson) + query(md + 1, r, rson);}    int C = ~(1<<31), M = (1<<16)-1;int rnd(int last) {    a = (36969 + (last >> 3)) * (a & M) + (a >> 16);    b = (18000 + (last >> 3)) * (b & M) + (b >> 16);    return (C & ((a << 16) + b)) % 1000000000;}int main(){    int cas;    scanf("%d", &cas);    while(cas--){        scanf("%d%d%d%d", &n, &m, &a, &b);        for(int i = 1; i <= n; ++i)             scanf("%d", &A[i]);        for(int i = 1; i <= n; ++i)            scanf("%d", &B[i]);        tot = 0;        build(1, n, 1);        int last = 0, ans = 0;        for(int i = 1; i <= m; ++i){            int l = rnd(last) % n + 1, r = rnd(last) % n + 1, x = rnd(last) + 1;            if(l > r)  swap(l, r);            if((l + r + x) & 1){                x = upper_bound(san + st[1], san + ed[1], x) - san;                update(l, r, x - 1, 1, n, 1);            }            else{                last = query(l, r, 1, n, 1);                ans += 1LL * i * last % mod;                if(ans >= mod) ans -= mod;            }        }        printf("%d\n", ans);    }    return 0;}