Description
Input
Output
Sample Input
1 2 3 4 5
A 1 5 4
M 3 5 1
A 1 5 4
Sample Output
3
HINT
先分块,用a数组存原数,b数组存每个块内排序好的数
修改的时候,整块直接打标记,小块的话就在a数组内暴力修改,并将暴力修改的地方更新到b数组中并排好序
查询的时候,对两端暴力统计。中间的由于已经排序完,所以可以二分
复杂度O(懒得算了反正能过)
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 #include<algorithm> 6 #define N (1000000+100) 7 using namespace std; 8 int Add[N],unit,num; 9 int n,m,a[N],b[N]; 10 int L[N],R[N],ID[N]; 11 void Build() 12 { 13 unit=sqrt(n); 14 num=n/unit; 15 if (n%unit) num++; 16 for (int i=1;i<=num;++i) 17 L[i]=(i-1)*unit+1,R[i]=i*unit; 18 R[num]=n; 19 for (int i=1;i<=n;++i) 20 ID[i]=(i-1)/unit+1; 21 for (int i=1;i<=num;++i) 22 sort(b+L[i],b+R[i]+1); 23 } 24 25 void Resort(int x) 26 { 27 for (int i=L[x];i<=R[x];++i) b[i]=a[i]; 28 sort(b+L[x],b+R[x]+1); 29 } 30 31 void Update(int l,int r,int k) 32 { 33 if (ID[l]==ID[r]) 34 { 35 for (int i=l;i<=r;++i) a[i]+=k; 36 Resort(ID[l]); 37 return; 38 } 39 for (int i=l;i<=R[ID[l]];++i) a[i]+=k; 40 for (int i=L[ID[r]];i<=r;++i) a[i]+=k; 41 Resort(ID[l]);Resort(ID[r]); 42 for (int i=ID[l]+1;i<=ID[r]-1;++i) Add[i]+=k; 43 } 44 45 int Query(int l,int r,int k) 46 { 47 int ans=0; 48 if (ID[l]==ID[r]) 49 { 50 for (int i=l;i<=r;++i) if (a[i]+Add[ID[i]]>=k) ans++; 51 return ans; 52 } 53 for (int i=l;i<=R[ID[l]];++i) if (a[i]+Add[ID[i]]>=k) ans++; 54 for (int i=L[ID[r]];i<=r;++i) if (a[i]+Add[ID[i]]>=k) ans++; 55 for (int i=ID[l]+1;i<=ID[r]-1;++i) 56 { 57 int p=lower_bound(b+L[i],b+R[i]+1,k-Add[i])-b; 58 ans+=R[i]-p+1; 59 } 60 return ans; 61 } 62 63 int main() 64 { 65 int l,r,k; 66 char p; 67 scanf("%d%d",&n,&m); 68 for (int i=1;i<=n;++i) 69 scanf("%d",&a[i]),b[i]=a[i]; 70 Build(); 71 for (int i=1;i<=m;++i) 72 { 73 scanf("\n%c%d%d%d",&p,&l,&r,&k); 74 if (p=='M') Update(l,r,k); 75 if (p=='A') printf("%d\n",Query(l,r,k)); 76 } 77 }