Cf52C
读完题你会发现就是个线段树区间加 区间求最小值
因为是循环的 可能x > y 代表 x - n,和 1 - y 区间
只不过读入数据要注意一点 这样的话我们读完两个数要读一个字符
如果这个字符不是回车 那么需要再读入一个
/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
ll Min(ll a,ll b){if(a<b) return a;return b;}
const int MAX_N = 200025;
ll arr[MAX_N];
namespace sgt
{
#define mid ((l+r)>>1)
ll minn[MAX_N<<2],col[MAX_N<<2];
void up(int rt) {minn[rt] = Min(minn[rt<<1],minn[rt<<1|1]);}
void pushnow(int rt,ll v) {minn[rt] += v;col[rt]+=v;}
void down(int rt) {if(col[rt]){pushnow(rt<<1,col[rt]),pushnow(rt<<1|1,col[rt]),col[rt] = 0;}}
void build(int rt,int l,int r){if(l==r) {minn[rt] = arr[l];return ;}build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);up(rt);}
void update(int rt,int l,int r,int x,int y,int v){
if(x<=l&&r<=y) return pushnow(rt,v);
down(rt);
if(x>mid) update(rt<<1|1,mid+1,r,x,y,v);
else if(y<=mid) update(rt<<1,l,mid,x,y,v);
else update(rt<<1,l,mid,x,y,v),update(rt<<1|1,mid+1,r,x,y,v);
up(rt);
}
ll query(int rt,int l,int r,int x,int y)
{
if(x<=l&&r<=y) return minn[rt];
down(rt);
if(x>mid) return query(rt<<1|1,mid+1,r,x,y);
else if(y<=mid) return query(rt<<1,l,mid,x,y);
else return Min(query(rt<<1,l,mid,x,y),query(rt<<1|1,mid+1,r,x,y));
}
#undef mid
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
int x,y,Q,n;scanf("%d",&n);
ll v;
for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]);
sgt::build(1,1,n);
scanf("%d",&Q);
while(Q--)
{
bool flag = false;
scanf("%d%d",&x,&y);
x++,y++;
if(x>y) flag = true;
char ch = getchar();
if(ch!='\n')
{
scanf("%lld",&v);
if(flag)
{
sgt::update(1,1,n,x,n,v);
sgt::update(1,1,n,1,y,v);
}
else
{
sgt::update(1,1,n,x,y,v);
}
}
else
{
if(flag)
{
printf("%lld\n",Min(sgt::query(1,1,n,x,n),sgt::query(1,1,n,1,y)));
}
else
{
printf("%lld\n",sgt::query(1,1,n,x,y));
}
}
}
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}