Codeforces 1053 C - Putting Boxes Together

C - Putting Boxes Together

思路：

求带权中位数

用树状数组维护修改

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair
#define pli pair
#define pii pair
#define piii pair
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 3e5 + 10;
const int MOD = 1e9 + 7;
int a[N], w[N], n;
struct BIT {
    LL bit[N];
    int ty;
    void add(int x, LL a) {
        while(x <= n) {
            if(ty == 0) bit[x] = (bit[x] + a) % MOD;
            else bit[x] = bit[x] + a;
            x += x&-x;
        }
    }
    LL sum(int x) {
        LL res = 0;
        while(x) {
            if(ty == 0)res = (res + bit[x]) % MOD;
            else res = res + bit[x];
            x -= x&-x;
        }
        return res;
    }
}b1, b2;

void solve (int x, int y) {
    if(x == y) {
        printf("0\n");
        return ;
    }
    int l = x, r = y, m = l+r >> 1;
    LL tot = b1.sum(r) - b1.sum(l-1);
    LL sub = b1.sum(l-1);
    while(l < r) {
        if((b1.sum(m) - sub)*2 >= tot) r = m;
        else l = m+1;
        m = l+r >> 1;
    }
    LL ans = 0, cnt = a[m] - m;
    cnt %= MOD;
    ans = (ans + cnt * ((b1.sum(m) - b1.sum(x-1)) % MOD) % MOD - (b2.sum(m) - b2.sum(x-1))) % MOD;
    ans = (ans - cnt * ((b1.sum(y) - b1.sum(m)) % MOD) % MOD + (b2.sum(y) - b2.sum(m))) % MOD;
    ans = (ans + MOD) % MOD;
    printf("%lld\n", ans);

}
int main() {
    int q, x, y;
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
    b1.ty = 1;
    b2.ty = 0;
    for (int i = 1; i <= n; i++) {
        b1.add(i, w[i]);
        b2.add(i, 1LL*w[i]*(a[i]-i));
    }
    while(q--) {
        scanf("%d %d", &x, &y);
        if(x < 0) {
            x = -x;
            b1.add(x, -w[x]);
            b2.add(x, -1LL*w[x]*(a[x]-x));

            w[x] = y;
            b1.add(x, w[x]);
            b2.add(x, 1LL*w[x]*(a[x]-x));
        }
        else solve(x, y);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/widsom/p/9742581.html