• poj A Simple Problem with Integers


    A Simple Problem with Integers
    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 37225   Accepted: 10745
    Case Time Limit: 2000MS

    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

    Hint

    The sums may exceed the range of 32-bit integers.

    Source

     
    分析:线段树:成段增加,成段求和。
    #include<cstdio>
    #include<cstring>
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    typedef long long u64;
    u64 sum[1 << 18];
    int col[1 << 18];
    
    void pushup(int rt) {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    
    void pushdown(int rt, int len) {
        if (col[rt]) {
            col[rt << 1] += col[rt];
            col[rt << 1 | 1] += col[rt];
            sum[rt << 1] += (u64) (len - (len >> 1)) * col[rt];
            sum[rt << 1 | 1] += (u64) (len >> 1) * col[rt];
            col[rt] = 0;
        }
    }
    
    void build(int l, int r, int rt) {
        if (l == r) {
            scanf("%lld", &sum[rt]);
            return;
        }
        int m = (l + r) >> 1;
        build(lson);
        build(rson);
        pushup(rt);
    }
    
    void update(int L, int R, int c, int l, int r, int rt) {
        if (L <= l && R >= r) {
            col[rt] += c;
            sum[rt] += (u64) c * (r - l + 1);
            return;
        }
        pushdown(rt, r - l + 1);
        int m = (l + r) >> 1;
        if (L <= m)
            update(L, R, c, lson);
        if (R > m)
            update(L, R, c, rson);
        pushup(rt);
    }
    
    u64 query(int L, int R, int l, int r, int rt) {
        if (L <= l && R >= r) {
            return sum[rt];
        }
        pushdown(rt, r - l + 1);
        int m = (l + r) >> 1;
        u64 ans = 0;
        if (L <= m)
            ans += query(L, R, lson);
        if (R > m)
            ans += query(L, R, rson);
        return ans;
    }
    
    int main() {
        int n, m, a, b, c;
        char op[4];
        while (scanf("%d%d", &n, &m) != EOF) {
            memset(col, 0, sizeof (col));
            build(1, n, 1);
            while (m--) {
                scanf("%s", op);
                if (op[0] == 'Q') {
                    scanf("%d%d", &a, &b);
                    printf("%lld\n", query(a, b, 1, n, 1));
                } else if (op[0] == 'C') {
                    scanf("%d%d%d", &a, &b, &c);
                    update(a, b, c, 1, n, 1);
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/baidongtan/p/2717023.html
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