Codeforces Round #306 (Div. 2) B.Preparing Olympiad (位运算)
2016-04-12 22:58
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B. Preparing Olympiad
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You have n problems. You have estimated the difficulty of the i-th
one as integer ci. Now you
want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and
at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ 106)
— the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106)
— the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
input
output
input
output
input
output
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
题意:
给出n个问题,每个问题有自己的难度。要求你选出一些问题,这些问题的难度和要在l与r之间,且这些问题中最难的和最简单的难度差至少达到x,问你能有多少种可以选?
分析:
之前还算做过位运算的题,这题根本懵逼,太小看B题了,受到别人用位运算启发,由于n范围极小,完全可以通过枚举所有状态暴力计算,最后输出结果。
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You have n problems. You have estimated the difficulty of the i-th
one as integer ci. Now you
want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and
at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ 106)
— the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106)
— the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
input
3 5 6 1 1 2 3
output
2
input
4 40 50 10 10 20 30 25
output
2
input
5 25 35 10 10 10 20 10 20
output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
题意:
给出n个问题,每个问题有自己的难度。要求你选出一些问题,这些问题的难度和要在l与r之间,且这些问题中最难的和最简单的难度差至少达到x,问你能有多少种可以选?
分析:
之前还算做过位运算的题,这题根本懵逼,太小看B题了,受到别人用位运算启发,由于n范围极小,完全可以通过枚举所有状态暴力计算,最后输出结果。
#include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cstdlib> #include <string> #include <vector> #include <cstdio> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> using namespace std; #define INF 0x3f3f3f3f const int N=100005; const int mod=1e9+7; int n,l,r,x,c[16]; int main() { while (cin>>n>>l>>r>>x) { for (int i=0; i<n; i++) { cin>>c[i]; } int total=1<<n,sig=0,sum,num,maxx,minn; for (int i=0; i<total; i++) { sum=0,maxx=0,minn=INF,num=0; for (int j=0; j<n; j++) { if (i&(1<<j)) { sum+=c[j]; num++; maxx=max(maxx,c[j]); minn=min(minn,c[j]); } } if (num<2||sum>r||sum<l||maxx-minn<x) { continue; } sig++; } cout<<sig<<endl; } return 0; }
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