Codeforces 52C (线段树区间更新)

C. Circular RMQ

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:

• inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
• rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).

Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.

Write program to process given sequence of operations.

Input

The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), aiare integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v(0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.

Output

For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample test(s)

input

41 2 3 443 03 0 -10 12 1

output

100

题目链接：http://codeforces.com/problemset/problem/52/C

题目大意：求环型数组的rmq

题目分析：裸的区间更新

#include <cstdio>#include <cstring>#include <algorithm>#define ll long long#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1using namespace std;int const MAX = 200005;int const INF = 0x3fffffff;char s[100];int mi[MAX << 2], lazy[MAX << 2];int n, q;void PushUp(int rt){mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]);}void PushDown(int rt){if(lazy[rt]){mi[rt << 1] += lazy[rt];mi[rt << 1 | 1] += lazy[rt];lazy[rt << 1] += lazy[rt];lazy[rt << 1 | 1] += lazy[rt];lazy[rt] = 0;}}void Build(int l, int r, int rt){lazy[rt] = 0;if(l == r){scanf("%d", &mi[rt]);return;}int mid = (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){mi[rt] += c;lazy[rt] += c;return;}int mid = (l + r) >> 1;PushDown(rt);if(L <= mid)Update(L, R, c, lson);if(mid < R)Update(L, R, c, rson);PushUp(rt);}int Query(int L, int R, int l, int r, int rt){if(L <= l && r <= R)return mi[rt];int mid = (l + r) >> 1;int ans = INF;PushDown(rt);if(L <= mid)ans = min(ans, Query(L, R, lson));if(mid < R)ans = min(ans, Query(L, R, rson));return ans;}int main(){scanf("%d", &n);Build(1, n, 1);scanf("%d", &q);while(q --){int l, r, c;scanf("%d %d", &l, &r);l ++;r ++;if(getchar() == ' '){scanf("%d", &c);if(l <= r)Update(l, r, c, 1, n, 1);else{Update(1, r, c, 1, n, 1);Update(l, n, c, 1, n, 1);}}else{if(l <= r)printf("%d\n", Query(l, r, 1, n, 1));elseprintf("%d\n", min(Query(1, r, 1, n, 1), Query(l, n, 1, n, 1)));}}}