poj 3468 A Simple Problem with Integers || 线段树
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poj 3468 A Simple Problem with Integers 请戳
题意:
给 n 个数,对这组数用两种方法蹂躏:
* 一种蹂躏方法是给一段区间加上或者减去一个数;
* 一种是给出一段区间的和;
好好玩的样子。。。蹂躏起来思路:
很明显线段数蹂躏大法好,区间修改与查询。小傻子写了一个非主流版本,虽然ac了,但是时间用的还是比较长的。因为只有pushup,而没有pushdown。复杂度:
时间复杂度:O(n)=nlogn
空间复杂度:O(n)=8n 代码:
/* ***********************************************Author :IlovezilianCreated Time :2015/9/5 0:11:25File Name :seg_tree_5.cpp************************************************ *///#include <bits/stdc++.h>#include<cstdio>#include<algorithm>#include<cstring>#define INF 0x7fffffff#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const int N = 1e5 + 10, mod = 1e9+7;typedef long long LL;LL po[N<<2], add[N<<2];void pushup(int l, int r, int rt){ po[rt] = po[rt<<1] + po[rt<<1|1] + add[rt] * (r - l + 1); ///这里运算重复太多次}void build(int l, int r, int rt){ add[rt] = 0; if(l == r) { scanf("%I64d", po + rt); return; } int m = (l + r) >> 1; build(lson); build(rson); pushup(l, r, rt);}void update(int val, int L, int R, int l, int r, int rt){ if(L <= l && r <= R) { add[rt] += val; po[rt] += (r - l + 1) * val; return; } int m = (l + r) >> 1; if(L <= m) update(val, L, R, lson); if(R > m) update(val, L, R, rson); pushup(l, r, rt);}LL query(int L, int R, int l, int r, int rt){ if(L <= l && r <= R) { //printf("po(%d,%d) = %I64d\n", l, r, po[rt]); return po[rt]; } int m = (l + r) >> 1; LL ret = 0; if(L <= m) ret += query(L, R, lson) + add[rt] * (min(m - max(l, L) + 1, R - max(l, L) + 1)); if(R > m) ret += query(L, R, rson) + add[rt] * (min(min(R, r) - m, min(R, r) - L + 1)); //printf("po(%d,%d) = %I64d ret = %I64d\n", l, r, po[rt], ret); return ret;}void solve(){ int n, q; while(~scanf("%d%d", &n, &q)) { build(1, n, 1); char s[3]; int a, b, c; for(int i = 0; i < q; i ++) { scanf("%s", s); if(s[0] == 'C') { scanf("%d%d%d", &a, &b, &c); update(c, a, b, 1, n, 1); } else { scanf("%d%d", &a, &b); printf("%I64d\n", query(a, b, 1, n, 1)); } } }}int main(){ //freopen("","r",stdin); //freopen("","w",stdout); solve(); return 0;}
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