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  • POJ 3468 A Simple Problem with Integers(线段树 区间更新)

    Description

    You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

    Input

    The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
    The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
    Each of the next Q lines represents an operation.
    "C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
    "Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

    Output

    You need to answer all Q commands in order. One answer in a line.

    Sample Input

    10 5
    1 2 3 4 5 6 7 8 9 10
    Q 4 4
    Q 1 10
    Q 2 4
    C 3 6 3
    Q 2 4
    

    Sample Output

    4
    55
    9
    15

    线段树区间更新。


    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>
    #define lson o << 1, l, m
    #define rson o << 1|1, m+1, r
    using namespace std;
    typedef long long LL;
    const int MAX=0x3f3f3f3f;
    const int maxn = 111111;
    int n, q, a, b, c;
    LL sum[maxn<<2], add[maxn<<2];
    void up(int o) {
        sum[o] = sum[o<<1] + sum[o<<1|1];
    }
    void down(int o, int m) {
        if(add[o] != 0) {
            add[o<<1] += add[o];
            add[o<<1|1] += add[o];
            sum[o<<1] += add[o]*(m-(m>>1));
            sum[o<<1|1] += add[o]*(m>>1);
            add[o] = 0;
        }
    }
    void build(int o, int l, int r) {
        add[o] = 0;
        if(l == r) scanf("%I64d", &sum[o]);
        else {
            int m = (l+r) >> 1;
            build(lson);
            build(rson);
            up(o);
        }
    }
    LL query(int o, int l, int r) {
        if(a <= l && r <= b) return sum[o];
        down(o, r-l+1);
        int m = (l+r) >> 1;
        LL ans = 0;
        if(a <= m) ans += query(lson);
        if(m < b) ans += query(rson);
        return ans;
    }
    void update(int o, int l, int r) {
        if(a <= l && r <= b) {
            add[o] += c;
            sum[o] += (LL)c*(r-l+1);
            return;
        }
        int m = (l+r) >> 1;
        down(o, r-l+1);
        if(a <= m) update(lson);
        if(m < b) update(rson);
        up(o);
    }
    int main()
    {
        scanf("%d%d", &n, &q);
        build(1, 1, n);
        while(q--) {
            char op[3];
            scanf("%s",op);
            if(op[0] == 'C') {
                scanf("%d%d%d", &a, &b, &c);
                update(1, 1, n);
            } else {
                scanf("%d%d", &a, &b);
                printf("%I64d
    ", query(1, 1, n));
            }
        }
        return 0;
    }
    
    


    
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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/3983305.html
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