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Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 42670 | Accepted: 18577 |
Description
Input
Output
Sample Input
5
1 1
5 1
7 1
3 3
5 5
Sample Output
1
2
1
1
0
Hint
Source
1 /**/ 2 #include<iostream> 3 #include<cstdio> 4 #include<cmath> 5 #include<cstring> 6 #include<algorithm> 7 using namespace std; 8 const int mxn=50000; 9 long long t[mxn]; 10 int n,m,k; 11 int a[mxn]; 12 int r[mxn]; 13 inline int lowbit(int x){ 14 return x&-x; 15 } 16 void add(int p,int v){ 17 while(p<=mxn){ 18 t[p]+=v; 19 p+=lowbit(p); 20 } 21 return; 22 } 23 int sum(int p){ 24 int res=0; 25 while(p){ 26 res+=t[p]; 27 p-=lowbit(p); 28 } 29 return res; 30 } 31 int main(){ 32 scanf("%d",&n); 33 int i,j; 34 int x,y; 35 for(i=1;i<=n;i++){ 36 scanf("%d%d",&x,&y); 37 x++; 38 r[sum(x)]++; 39 add(x,1); 40 } 41 for(i=0;i<n;i++)printf("%d\n",r[i]); 42 return 0; 43 }
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原文地址:http://www.cnblogs.com/SilverNebula/p/5836963.html