P2801 教主的魔法（分块）

分块儿练习，维护每个区间有序，二分找大于它的即可

注意的是数组别开小，并且要建立两个数组用来处理边角问题

代码：

#include 
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;ib;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"< vei;
typedef vector vep;
typedef map mpii;
typedef map mpci;
typedef map mpsi;
typedef deque deqi;
typedef deque deqc;
typedef priority_queue mxpq;
typedef priority_queue,greater > mnpq;
typedef priority_queue mxpqii;
typedef priority_queue,greater > mnpqii;
const int maxn=1000005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return ab||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
	if(a==0) return b;
	if(b==0) return a;
	if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
	else if(!(b&1)) return gcd(a,b>>1);
	else if(!(a&1)) return gcd(a>>1,b);
	else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n,q,a[maxn],b[maxn],l,r,x;
char op[5];
int tot,id[maxn],from[maxn],to[maxn];
int add[maxn];

void pre(){
	re(i,1,n) a[i]=b[i];
	tot=sqrt(n);
	re(i,1,tot)
		from[i]=(i-1)*tot+1,
		to[i]=i*tot;
	if(to[tot]=x) ret++;
		return ret;
	}
	re(i,l,to[id[l]]) if(b[i]+add[id[i]]>=x) ret++;
	re(i,from[id[r]],r) if(b[i]+add[id[i]]>=x) ret++;
	re(i,id[l]+1,id[r]-1){
		int L=from[i],R=to[i];
		int p=lower_bound(a+L,a+R+1,x-add[i])-a;
//		cout<<"??? "<