• POJ-3468 A Simple Problem with Integers (区间求和,成段加减)


    You have N integers, A1A2, ... , 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 A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
    Each of the next Q lines represents an operation.
    "C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
    "Q a b" means querying the sum of AaAa+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

    区间求和,成段加减

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    
    using namespace std;
    #define lson l, m, rt<<1
    #define rson m+1, r, rt<<1|1
    typedef long long LL;
    const int N = 100000 + 5;
    LL T[N<<2],add[N<<2];
    
    void PushUP(int rt){
        T[rt] = T[rt<<1] + T[rt<<1|1];
    }
    
    void PushDown(int rt, int m){
        if(add[rt]){
            add[rt<<1] += add[rt];
            add[rt<<1|1] += add[rt];
            T[rt<<1] += add[rt]*(LL)(m - (m >> 1));
            T[rt<<1|1] += add[rt]*(LL)(m >> 1);
            add[rt] = 0;
        }
    }
    void Build(int l, int r, int rt){
        add[rt] = 0;
        if(l == r){
            scanf("%lld",&T[rt]);
            return ;
        }
        int m = (l + r) >> 1;
        Build(lson);
        Build(rson);
        PushUP(rt);
    }
    
    void Updata(int L,int R, int c,int l, int r, int rt){
        if(L <= l && r <= R){
            add[rt] += c;
            T[rt] += (LL)c*(r - l + 1);
            return ;
        }
        PushDown(rt, r - l + 1);
        int m = (l + r) >> 1;
        if(L <= m) Updata(L, R, c, lson);
        if(R > m) Updata(L, R, c, rson);
        PushUP(rt);
    }
    
    LL Query(int L, int R, int l, int r, int rt){
        if(L <= l && r <= R) return T[rt];
        PushDown(rt, r - l + 1);
        int m  = (l + r) >> 1;
        LL ret = 0;
        if(L <= m) ret += Query(L, R, lson);
        if(R > m) ret += Query(L, R, rson);
        return ret;
    }
    
    int main(){
        int n,m;
        while(scanf("%d %d",&n,&m)==2){
            Build(1, n, 1);
            char ch[5];
            int a,b,c;
            while(m--){
                scanf("%s %d %d",ch,&a,&b);
                if(ch[0] == 'Q')printf("%lld
    ",Query(a, b, 1, n, 1));
                else{
                    scanf("%d",&c);
                    Updata(a, b, c, 1, n, 1);
                }
            }
        }
        return 0;
    }
     


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  • 原文地址:https://www.cnblogs.com/Pretty9/p/7384046.html
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