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  • POJ3468A Simple Problem with Integers(线段树区间修改)

    http://www.notonlysuccess.com/index.php/segment-tree-complete/comment-page-2/#comment-4139

    题意:给一组数据,有两个询问,一种是给区间[a,b]上每个数加c,另一种是求区间[a,b]的和!

    #include <cstdio>
    #include <algorithm>
    using namespace std;
     
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    #define LL long long
    const int maxn = 111111;
    LL add[maxn<<2];
    LL sum[maxn<<2];
    void PushUp(int rt) {
        sum[rt] = sum[rt<<1] + sum[rt<<1|1];
    }
    void PushDown(int rt,int m) {
        if (add[rt]) {
            add[rt<<1] += add[rt];
            add[rt<<1|1] += add[rt];
            sum[rt<<1] += add[rt] * (m - (m >> 1));
            sum[rt<<1|1] += add[rt] * (m >> 1);
            add[rt] = 0;
        }
    }
    void build(int l,int r,int rt) {
        add[rt] = 0;
        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) {
            add[rt] += c;
            sum[rt] += (LL)c * (r - l + 1);
            return ;
        }
        PushDown(rt , r - l + 1);
        int m = (l + r) >> 1;
        if (L <= m) update(L , R , c , lson);
        if (m < R) update(L , R , c , rson);
        PushUp(rt);
    }
    LL 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;
        LL ret = 0;
        if (L <= m) ret += query(L , R , lson);
        if (m < R) ret += query(L , R , rson);
        return ret;
    }
    int main() {
        int N , Q;
        scanf("%d%d",&N,&Q);
        build(1 , N , 1);
        while (Q --) {
            char op[2];
            int a , b , c;
            scanf("%s",op);
            if (op[0] == 'Q') {
                scanf("%d%d",&a,&b);
                printf("%lld\n",query(a , b , 1 , N , 1));
            } else {
                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/arbitrary/p/3054485.html
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