• Codeforces Round #179 (Div. 1) A. Greg and Array 离线区间修改


    A. Greg and Array

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/contest/295/problem/A

    Description

    Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: liridi(1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

    Greg wrote down k queries on a piece of paper. Each query has the following form: xiyi(1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

    Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

    Input

    The first line contains integers nmk (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

    Next m lines contain operations, the operation number i is written as three integers: liridi(1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

    Next k lines contain the queries, the query number i is written as two integers: xiyi(1 ≤ xi ≤ yi ≤ m).

    The numbers in the lines are separated by single spaces.

    Output

    On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

    Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64dspecifier.

    Sample Input

    3 3 3
    1 2 3
    1 2 1
    1 3 2
    2 3 4
    1 2
    1 3
    2 3

    Sample Output

    9 18 17

    HINT

    题意

    给你n个数,然后有m个命令,每个命令是将[l,r]中的数都增加d,有k次真正要执行的操作(。。),是将第[l,r]的命令执行一遍

    然后问你,这个n个数,最后悔变成什么模样。

    题解:

    这道题扫一遍就好了,由于只有最后一次询问,那么我们可以离线去做,而不用去写什么线段树

    首先我们离线处理操作,然后得到每一个命令都会执行多少次,然后我们再离线去处理命令就好了

    离线处理,就是在起始端和结束段打上标记,然后扫一遍就好了

    代码:

    #include<iostream>
    #include<stdio.h>
    using namespace std;
    #define maxn 100005
    long long a[maxn];
    long long flag[maxn];
    long long t[maxn];
    struct OP
    {
        int x,y;
        long long val;
    };
    OP op[maxn];
    int main()
    {
        int n,m,k;scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        for(int i=1;i<=m;i++)
        {
            int x,y;long long z;
            scanf("%d%d%lld",&op[i].x,&op[i].y,&op[i].val);
        }
        for(int i=1;i<=k;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            t[x]++;
            t[y+1]--;
        }
        long long sum = 0;
        for(int i=1;i<=m;i++)
        {
            sum+=t[i];
            op[i].val*=sum;
        }
        for(int i=1;i<=m;i++)
        {
            flag[op[i].x]+=op[i].val;
            flag[op[i].y+1]-=op[i].val;
        }
        sum = 0;
        for(int i=1;i<=n;i++)
        {
            sum+=flag[i];
            printf("%lld ",a[i]+sum);
        }
        printf("
    ");
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5015921.html
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