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  • Codeforces Round #470 (rated, Div. 2, based on VK Cup 2018 Round 1)C. Producing Snow

    http://codeforces.com/contest/948/problem/C

    C. Producing Snow

    Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.

    Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.

    Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.

    You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.

    Input

    The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.

    The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.

    The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.

    Output

    Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.

    Examples
    input
    Copy
    3
    10 10 5
    5 7 2
    output
    5 12 4
    input
    Copy
    5
    30 25 20 15 10
    9 10 12 4 13
    output
    9 20 35 11 25
    Note

    In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.

     题意是给定长度n,第一行第i项是第i天造出来的雪的体积Vi,第二行第i项是第i天气温能够融化掉雪的体积Ti,求第i天当天总共融了多少雪。因此对Ti求前缀和可以得到总共融雪体积,对于每一项Vi可以二分前缀和数组得到在第Q天,这堆雪全部融化完,即在[i,Q-1]范围内这堆雪每天融化Ti,用树状数组对它进行区间更新;第Q天融掉这堆雪剩下的量,直接计入答案数组。最后只需要在答案数组中将树状数组单点的值和当天的Ti相乘加进去就可以了。

    有两个地方要注意,一个是如果查询到的Q是最后一天,需要特判融化的量应该是Ti和雪剩余量中的较小值;还有一个是查询到的Q如果比当前天i要小不计入计算。

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <vector>
    #include <queue>
    #include <stack>
    #include <set>
    #include <map>
    #define INF 0x3f3f3f3f
    #define lowbit(x) (x&(-x))
    #define eps 0.00000001
    #define pn printf("
    ")
    using namespace std;
    typedef long long ll;
    
    const int maxn = 1e6+5;
    int n;
    ll v[maxn], t[maxn], to[maxn];
    ll ans[maxn];
    ll c[maxn];
    
    void update(int pos,int val)
    {
        while(pos <= maxn)
        {
            c[pos] += val;
            pos += lowbit(pos);
        }
    }
    int query(int pos)
    {
        int ret = 0;
        while(pos > 0)
        {
            ret += c[pos];
            pos -= lowbit(pos);
        }
        return ret;
    }
    
    int lower(ll x)
    {
        int l = 1, r = n, mid;
        while(l < r)
        {
            mid = (l + r) >> 1;
            if(to[mid] >= x) r = mid;
            else l = mid + 1;
        }
        return l;
    }
    
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%I64d", v+i);
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d", t+i);
            to[i] = to[i-1] + t[i];
        }
        for(int i=1;i<=n;i++)
        {
            ll tp = v[i] + to[i-1];
            int pos = lower(tp);
            if(i < pos)
            {
                update(i, 1);
                update(pos, -1);
            }
            if(i <= pos)
            {
                if(tp > to[pos]) ans[pos] += t[pos];
                else ans[pos] += v[i] - (to[pos-1] - to[i-1]);
            }
        }
        for(int i=1;i<=n;i++)
            ans[i] += 1LL * t[i] * query(i);
        for(int i=1;i<=n;i++)
        {
            if(i != 1) printf(" ");
            printf("%I64d", ans[i]);
        } pn;
    }
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  • 原文地址:https://www.cnblogs.com/HazelNut/p/8543562.html
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