本文主要是介绍BestCoder Round #45 (1,2)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
比赛链接：http://bestcoder.hdu.edu.cn/contests/contest_show.php?cid=604
Dylans loves numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 94 Accepted Submission(s): 67
Total Submission(s): 94 Accepted Submission(s): 67
Problem Description
Who is Dylans?You can find his ID in UOJ and Codeforces.
His another ID is s1451900 in BestCoder.
And now today's problems are all about him.
Dylans is given a number N .
He wants to find out how many groups of "1" in its Binary representation.
If there are some "0"（at least one）that are between two "1",
then we call these two "1" are not in a group,otherwise they are in a group.
His another ID is s1451900 in BestCoder.
And now today's problems are all about him.
Dylans is given a number N .
He wants to find out how many groups of "1" in its Binary representation.
If there are some "0"（at least one）that are between two "1",
then we call these two "1" are not in a group,otherwise they are in a group.
In the first line there is a number T .
T is the test number.
In the next T lines there is a number N .
0≤N≤1018,T≤1000
T is the test number.
In the next T lines there is a number N .
0≤N≤1018,T≤1000
For each test case,output an answer.
1 5
2
BestCoder Round #45
题目大意：求一个数字二进制中多少个连续的1群体
题目分析：直接模拟
题目大意：求一个数字二进制中多少个连续的1群体
题目分析：直接模拟
#include <cstdio>
#define ll long longint main()
{int T;scanf("%d", &T);while(T--){ll n;scanf("%I64d", &n);int cnt = 0;bool flag = true;while(n != 0){if((n & 1) ^ 0){if(flag)cnt ++;flag = false;}elseflag = true;n >>= 1;}printf("%d\n", cnt);}
}Dylans loves sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 195 Accepted Submission(s): 105
Total Submission(s): 195 Accepted Submission(s): 105
Problem Description
Dylans is given N numbers a[1]....a[N]
And there are Q questions.
Each question is like this (L,R)
his goal is to find the “inversions” from number L to number R.
more formally,his needs to find the numbers of pair（x,y）,
that L≤x,y≤R and x<y and a[x]>a[y]
And there are Q questions.
Each question is like this (L,R)
his goal is to find the “inversions” from number L to number R.
more formally,his needs to find the numbers of pair（x,y）,
that L≤x,y≤R and x<y and a[x]>a[y]
In the first line there is two numbers N and Q.
Then in the second line there are N numbers:a[1]..a[N]
In the next Q lines,there are two numbers L,R in each line.
N≤1000,Q≤100000,L≤R,1≤a[i]≤231−1
Then in the second line there are N numbers:a[1]..a[N]
In the next Q lines,there are two numbers L,R in each line.
N≤1000,Q≤100000,L≤R,1≤a[i]≤231−1
For each query,print the numbers of "inversions”
3 2 3 2 1 1 2 1 3
1 3HintYou shouldn't print any space in each end of the line in the hack data.
Source
BestCoder Round #45
题目分析：用dp[i][j]预处理，先算区间[i,j]中与a[i]成逆序对的数目，然后从后向前累加，相当于加一个a[j]，看它对整个区间的逆序数对数是否有影响
#include <cstdio>
#include <cstring>
int const MAX = 1005;
int dp[MAX][MAX];
int a[MAX];int main()
{int n, q;scanf("%d %d", &n, &q);for(int i = 1; i <= n; i++)scanf("%d", &a[i]);memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; i++)for(int j = i; j <= n; j++)dp[i][j] += (dp[i][j - 1] + (a[i] > a[j] ? 1 : 0));for(int j = n; j > 0; j--)for(int i = j; i > 0; i--)dp[i][j] += dp[i + 1][j];while(q--){int l, r;scanf("%d %d", &l, &r);printf("%d\n", dp[l][r]);}
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