CodeForces 52C Circular RMQ（区间循环线段树，区间更新，区间求和）

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 test(s)

input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1

output
1
0
0

#include <cstdio> #include <algorithm> using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 //lson和rson分辨表示结点的左儿子和右儿子 //rt表示当前子树的根(root),也就是当前所在的结点 #define INF 0x3fffffff __int64 mmin[5000047]; __int64 col[5000047]; __int64 MIN(__int64 a, __int64 b) { if( a < b) return a; return b; } void pushup(int rt) { //sum[rt] = sum[rt<<1] + sum[rt<<1|1]; mmin[rt]=MIN(mmin[rt<<1], mmin[rt<<1|1]); } void pushdown(int rt, int m) { if(col[rt]) { col[rt<<1]+=col[rt]; col[rt<<1|1]+=col[rt]; mmin[rt<<1]+=col[rt]; mmin[rt<<1|1]+=col[rt]; col[rt]=0; } } void build(int l, int r, int rt) { col[rt]=0; if(l==r) { scanf("%I64d", &mmin[rt]); return ; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt); } void update(int L, int R, int v, int l, int r, int rt) { if(L<=l && r<=R) { col[rt]+=v; mmin[rt]+=v; return ; } pushdown(rt, r-l+1); int m=(l+r)>>1; if(L<=m) update(L, R, v, lson); if(R>m) update(L, R, v, rson); pushup(rt); } __int64 query(int L, int R, int l, int r, int rt) {//查询区间[L，R]中的最小值 if(L<=l && r<=R)//当前结点完全包含在查询区间内 return mmin[rt]; pushdown(rt, r-l+1); int m=(l+r)>>1; __int64 ret=INF; if(L<=m)//往左走 ret=MIN(ret, query(L, R, lson)); if(R>m)//往右走 ret=MIN(ret, query(L, R, rson)); return ret; } int main() { int n, m, a, b, c; char op; scanf("%d", &n); build(1, n, 1); scanf("%d", &m); while(m--) { scanf("%d%d%c", &a, &b, &op); a++;//将区间转换为是1到n b++; if(a<=b) { if(op==' ')//意味着是输入三个数的操作 { scanf("%d", &c); if(c==0) continue; update(a, b, c, 1, n, 1); } else printf("%I64d\n", query(a, b, 1, n, 1)); } else { if(op==' ') { scanf("%d", &c); if(c==0) continue; update(a, n, c, 1, n, 1);//拆分区间为a到n和1到b update(1, b, c, 1, n, 1); } else printf("%I64d\n", MIN(query(a, n, 1, n, 1), query(1, b, 1, n, 1))); } } return 0; }