标签：efi 相同 family -- getc ini nts event tree
Given a binary number, we are about to do some operations on the number. Two types of operations can be here.
‘I i j‘ which means invert the bit from i to j (inclusive)
‘Q i‘ answer whether the ith bit is 0 or 1
The MSB (most significant bit) is the first bit (i.e. i=1). The binary number can contain leading zeroes.
Input
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each case starts with a line containing a binary integer having length n (1 ≤ n ≤ 105). The next line will contain an integerq (1 ≤ q ≤ 50000) denoting the number of queries. Each query will be either in the form ‘I i j‘ where i, j are integers and1 ≤ i ≤ j ≤ n. Or the query will be in the form ‘Q i‘ where i is an integer and 1 ≤ i ≤ n.
Output
For each case, print the case number in a single line. Then for each query ‘Q i‘ you have to print 1 or 0 depending on theith bit.
Sample Input
2
0011001100
6
I 1 10
I 2 7
Q 2
Q 1
Q 7
Q 5
1011110111
6
I 1 10
I 2 7
Q 2
Q 1
Q 7
Q 5
Sample Output
Case 1:
0
1
1
0
Case 2:
0
0
0
1
Solve:
区间问题,有查询有修改,所以直接BIT水过,每次查询的时候只要判断这个其和是不是奇数,奇数就取相反,偶数就相同
Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 static const int MAXN = 1e5 + 10; 4 #define CLR(x , v) memset(x , v , sizeof(x)); 5 #define CASE(x) printf("Case %d:\n", x); 6 char in[MAXN] = {0}; 7 int data[MAXN] = {0}; 8 int tree[MAXN << 1] = {0}; 9 int si = 0; 10 inline int lowbit(int x) {return x & (-x);} 11 void Update(int pos , int value) 12 { 13 for(int i = pos ; i <= si ; i += lowbit(i)) 14 { 15 tree[i] += value; 16 } 17 } 18 int Sum(int x) 19 { 20 int ans = 0; 21 for(int i = x ; i > 0 ; i -= lowbit(i)) 22 ans += tree[i]; 23 return ans; 24 } 25 int main() 26 { 27 int t , q , x , y; 28 char cmd; 29 scanf("%d" , &t); 30 for(int c = 1 ; c <= t ; ++c) 31 { 32 CLR(in , ‘\0‘); 33 CLR(tree , 0); 34 scanf(" %s" , in + 1); 35 CASE(c); 36 si = strlen(in + 1); 37 scanf("%d" , &q); 38 while(q--) 39 { 40 scanf(" %c%d" , &cmd , &x); 41 if(cmd == ‘I‘) 42 { 43 scanf("%d" , &y); 44 Update(x , 1); 45 Update(y + 1 , -1); 46 } 47 else 48 { 49 if(Sum(x) & 1) 50 { 51 printf("%d\n" , in[x] == ‘0‘ ? 1 : 0); 52 } 53 else 54 printf("%c\n" , in[x]); 55 } 56 } 57 58 } 59 return 0; 60 }
Given n points (1 dimensional) and q segments, you have to find the number of points that lie in each of the segments. A point pi will lie in a segment A B if A ≤ pi ≤ B.
For example if the points are 1, 4, 6, 8, 10. And the segment is 0 to 5. Then there are 2 points that lie in the segment.
Input
Input starts with an integer T (≤ 5), denoting the number of test cases.
Each case starts with a line containing two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers denoting the points in ascending order. All the integers are distinct and each of them range in [0, 108].
Each of the next q lines contains two integers Ak Bk (0 ≤ Ak ≤ Bk ≤ 108) denoting a segment.
Output
For each case, print the case number in a single line. Then for each segment, print the number of points that lie in that segment.
Sample Input
1
5 3
1 4 6 8 10
0 5
6 10
7 100000
Sample Output
Case 1:
2
3
2
Means:
给你一些点,然后给你一条线段,然后问你有多少点在线段上
Solve:
简单二分查找,因为给出的点都是有序的,所以直接存点就行了,二分找第一个大于等于线段左侧端点的点,找最后一个小于等于线段右侧端点的点,之间的数量就是了
Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LSON id << 1 , l , mid 4 #define RSON id << 1 | 1 , mid + 1 , r 5 #define ROOT 1 , 0 , n 6 #define CLR(x , y) memset(x , y , sizeof(x)) 7 #define LOWBIT(x) x & (-x) 8 #define FORN(i , a , n) for(int i = (a) ; i <= (n) ; ++i) 9 #define FORP(i , n , a) for(int i = (n) ; i >= (a) ; --i) 10 #define CASE(x) printf("Case %d:\n", x); 11 static const double EPS = 1e-8; 12 static const int INF = 0X3fffffff; 13 typedef long long LL; 14 int n , q; 15 static const int MAXN = 1e5 + 10; 16 LL data[MAXN]; 17 template<class T> inline 18 T read(T &x) 19 { 20 x = 0; 21 int f = 1 ; char ch = getchar(); 22 while (ch < ‘0‘ || ch > ‘9‘) {if (ch == ‘-‘) f = -1; ch = getchar();} 23 while (ch >= ‘0‘ && ch <= ‘9‘) {x = x * 10 + ch - ‘0‘; ch = getchar();} 24 x *= f; 25 } 26 int main() 27 { 28 int t; 29 read(t); 30 for(int c = 1 ; c <= t ; ++c) 31 { 32 int pos = 0; 33 CASE(c); 34 read(n) , read(q); 35 LL x; 36 for(int i = 0 ; i < n ; ++i) 37 { 38 read(x); 39 data[pos++] = x; 40 } 41 LL y; 42 while(q--) 43 { 44 read(x) , read(y); 45 LL temp1 = upper_bound(data , data + n , y) - data; 46 LL temp2 = lower_bound(data , data + n , x) - data; 47 printf("%lld\n" , temp1 - temp2); 48 } 49 } 50 }
标签：efi 相同 family -- getc ini nts event tree
原文地址：http://www.cnblogs.com/jianglingxin/p/6700714.html
