BestCoder Round #45 (1,2)

时间：2015-06-21 00:50:17 阅读：187 评论：0 收藏：0 [点我收藏+]

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

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.

Input

In the first line there is a number

T is the test number.

In the next

T lines there is a number

0≤N≤1018,T≤1000

Output

For each test case,output an answer.

Sample Input

1
5

Sample Output

Source

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

#include <cstdio>
#define ll long long

int 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;
            }
            else
                flag = 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

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\leqx,y\leqR

and

x<y

and

a[x]>a[y]

Input

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\leq1000,Q\leq100000,L\leqR,1\leqa[i]\leq231?1

Output

For each query,print the numbers of "inversions”

Sample Input

Sample Output

1
3

Hint

You shouldn‘t print any space in each end of the line in the hack data.

Source

BestCoder Round #45

题目大意：求区间逆序数对数

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]);
    }
}

BestCoder Round #45 (1,2)

标签：bestcoder

原文地址：http://blog.csdn.net/tc_to_top/article/details/46576591

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