poj 3468 A Simple Problem with Integers

来源：互联网发布：java target runtime 编辑：程序博客网时间：2024/05/29 19:54

原题链接：http://poj.org/showsource?solution_id=13997740
线段树，区间求和，成段更新，若有不明白的单步调式观察中间值，很好理解的。
不知道是我写的烂，还是线段树常数太大，反正程序跑的很慢o(╯□╰)o。
具体的看代码吧。

#include<stdio.h>#include<stdlib.h>#include<string.h>#define Max_N 200000typedef long long ll;typedef struct _seg{    ll val, addMark;}SegTree;SegTree seg[Max_N << 2];void built(ll root, ll x, ll y){    seg[root].addMark = 0;    if (x == y){        scanf("%lld", &seg[root].val);        return;    }    else {        ll mid = (x + y) >> 1;        built(root << 1, x, mid);        built(root << 1 | 1, mid + 1, y);        seg[root].val = seg[root << 1].val + seg[root << 1 | 1].val;    }}void pushDown(ll root, ll len){    if (seg[root].addMark != 0){        seg[root << 1].addMark += seg[root].addMark;        seg[root << 1 | 1].addMark += seg[root].addMark;        seg[root << 1].val += (len - (len >> 1)) * seg[root].addMark;        seg[root << 1 | 1].val += (len >> 1) * seg[root].addMark;        seg[root].addMark = 0;    }}void update(ll root, ll _x, ll _y, ll x, ll y, ll val){    if (x > _y || y < _x) return;    if (x <= _x  && _y <= y){        seg[root].addMark += val;        seg[root].val += val * (_y - _x + 1);        return;    }    pushDown(root, _y - _x + 1);    ll mid = (_x + _y) >> 1;    update(root << 1, _x, mid, x, y, val);    update(root << 1 | 1, mid + 1, _y, x, y, val);    seg[root].val = seg[root << 1].val + seg[root << 1 | 1].val;}ll query(ll root, ll _x, ll _y, ll x, ll y){    if (x > _y || y < _x) return 0;    if (x <= _x  && _y <= y) return seg[root].val;    pushDown(root, _y - _x + 1);    ll mid = (_x + _y) >> 1;    ll v1 = query(root << 1, _x, mid, x, y);    ll v2 = query(root << 1 | 1, mid + 1, _y, x, y);    return v1 + v2;}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);    freopen("out.txt", "w+", stdout);#endif    char ch;    ll a, b, c, N, Q;    while (~scanf("%lld %lld", &N, &Q)){        built(1, 1, N);        while (Q--){            getchar();            scanf("%c", &ch);            if ('Q' == ch){                scanf("%lld %lld", &a, &b);                printf("%lld\n", query(1, 1, N, a, b));            } else {                scanf("%lld %lld %lld", &a, &b, &c);                update(1, 1, N, a, b, c);            }        }    }    return 0;}

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