Description
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), ai are 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 Input
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
题意很好理解。
如果a>b的话，就查0~b和a~n-1，其余的就是线段树区间更新模板，判断m个询问里是否存在c，详见代码，很好理解。
#include
#include
#include
#include
#include
#include
#define lson o << 1, l, m #define rson o << 1|1, m+1, r using namespace std; typedef __int64 LL; const __int64 MAX = 9223372036854775807; const int maxn = 200010; int n, a, q, c, b; char str[1200]; LL mi[maxn<<2], add[maxn<<2]; void up(int o) { mi[o] = min(mi[o<<1], mi[o<<1|1]); } void down(int o) { if(add[o]) { add[o<<1] += add[o]; add[o<<1|1] += add[o]; mi[o<<1] += add[o]; mi[o<<1|1] += add[o]; add[o] = 0; } } void build(int o, int l, int r) { if(l == r) { scanf("%I64d", &mi[o]); return; } int m = (l+r) >> 1; build(lson); build(rson); up(o); } void update(int o, int l, int r) { if(a <= l && r <= b) { add[o] += c; mi[o] += c; return ; } down(o); int m = (l+r) >> 1; if(a <= m) update(lson); if(m < b ) update(rson); up(o); } LL query(int o, int l, int r) { if(a <= l && r <= b) return mi[o]; down(o); int m = (l+r) >> 1; LL res = MAX; if(a <= m) res = query(lson); if(m < b ) res = min(res, query(rson)); return res; } int main() { scanf("%d", &n); build(1, 0, n-1); scanf("%d", &q); getchar(); while(q--) { gets(str); if(sscanf(str,"%d %d %d", &a, &b, &c) == 3) { if(a <= b) update(1, 0, n-1); else { int tmp = b; b = n-1; update(1, 0, n-1); a = 0, b = tmp; update(1, 0, n-1); } } else { LL ans ; if(a <= b) ans = query(1, 0, n-1); else { int tmp = b; b = n-1; ans = query(1, 0, n-1); a = 0, b = tmp; ans = min(ans, query(1, 0, n-1)); } printf("%I64d\n", ans); } } return 0; }
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