Codeforces 706B Interesting drink
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Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It‘s known that the price of one bottle in the shop i is equal to xicoins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy‘s favourite drink.
The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
5
3 10 8 6 11
4
1
10
3
11
0
4
1
5
On the first day, Vasiliy won‘t be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
这题目可以,题目说xi不超过100000,于是我就开了个数组用树状数组,提交runtime error,一看,哦,查询的是范围是1000000000,所以加了一步判断,这次是答案错误,看样子xi也是这么大的范围,这样一来,这题怕是通解另有其他,估计是二分,可是树状数组不行了吗,数组开不了特别大,那就map一下,过了,再写个二分,二分耗时少,果然是二分。。。
树状数组代码:
#include <iostream> #include <map> #include <cstdio> #define MAX 1000000000 using namespace std; int n,m; map<int,int> sum; int lowbit(int t) { return t&(-t); } void update(int x) { while(x <= MAX) { sum[x] ++; x += lowbit(x); } } int getans(int x) { int ans = 0; while(x > 0) { ans += sum[x]; x -= lowbit(x); } return ans; } int main() { int d; scanf("%d",&n); for(int i = 0;i < n;i ++) { scanf("%d",&d); update(d); } scanf("%d",&m); for(int i = 0;i < m;i ++) { scanf("%d",&d); printf("%d ",getans(d)); } }
二分代码:
#include <iostream> #include <algorithm> #include <cstdio> #define MAX 100000 using namespace std; int n,m; int x[MAX]; int main() { int d; scanf("%d",&n); for(int i = 0;i < n;i ++) { scanf("%d",&x[i]); } sort(x,x + n); scanf("%d",&m); for(int i = 0;i < m;i ++) { scanf("%d",&d); int l = 0,r = n,mid; while(l < r) { mid = (l + r) / 2; if(x[mid] <= d) l = mid + 1; else r = mid; } printf("%d ",l); } }
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