2017 Multi-University Training Contest - Team 7

最新推荐文章于 2025-08-21 10:38:03 发布

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最新推荐文章于 2025-08-21 10:38:03 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/Suwako1997/p/7373981.html

本文精选了四个算法竞赛题目并提供了详细的解答思路，包括求模运算的可能性、构造Kolakoski序列、寻找最大权值和的直线以及多次执行前缀异或变换等。

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1005 Euler theorem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

HazelFan is given two positive integers

Input

The first line contains a positive integer

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input

Sample Output

题意：给定正整数a，对于任意正整数b，求a mod b有多少种可能的结果

题解：签到题，可知每个小于a/2的非负整数都可以成为模的结果，当b > a时，a mod b等于a本身，因此可以计算出结果。

include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int t,a;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&a);
        printf("%d\n",a - (a / 2) + 1);
    }
}

1011 Kolakoski

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

This is Kolakosiki sequence:

Input

The first line contains a positive integer

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input

Sample Output

题意：给你一个数列1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1……. ，求数列的第n项

补充：Kolakoski序列是一个仅由1和2组成的无限数列，是一种通过“自描述”来定义的数列。它的定义很简单，若把数列中相同的数定为一组，令a(1) = 1，a(2) = 2，则a(n)等于第n组数的长度。我们可以把数列看成1,22,11,2,1,22,1,22,11，……，因此可知a(1) = 1, a(2) = 2, a(3) = 2, ……

题解：数列构造题，根据数列的定义来构造数列即可，用两个指针扫一下。

#include<iostream>
#include<cstdio>
using namespace std;
int a[10000050];
int main()
{
    int i,j,temp,k,n,t;
    a[1] = 1;
    a[2] = 2;
    a[3] = 2;
    i = 3;j = 4;temp = 1;
    while (j <= 10000050)
    {
        for (k = 1;k <= a[i];k++)
        {
            a[j] = temp;
            j++;
        }
        i++;
        temp = 3 - temp;
    }
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        printf("%d\n",a[n]);
    }
}

1008 Hard challenge

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

There are

Input

The first line contains a positive integer

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input

1 1 1

1 -1 1

1 1 1

1 -1 10

-1 0 100

Sample Output

1100

题意：平面直角坐标系上有

题解：先按照每个点极角排序，按点在x轴的左边还是右边建立一个区间，之后用尺取法每次对区间挪动一点，找出最大值即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct node
{
    int x,y;
    double angle;
    long long val;
};
node a[500050];
int cmp(node x,node y)
{
    return x.angle < y.angle;
}
int main()
{
    int t,n,i;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        for (i = 1;i <= n;i++)
        {
            scanf("%d%d%lld",&a[i].x,&a[i].y,&a[i].val);
            if (a[i].x != 0) a[i].angle = atan(1.0 * a[i].y / a[i].x);
            else a[i].angle = 3.14159265 / 2;
        }
        sort(a + 1,a + n + 1,cmp);
        long long l = 0,r = 0;
        for (i = 1;i <= n;i++)
        {
            if (a[i].x >= 0) r = r + a[i].val;
            else l = l + a[i].val;
        }
        long long ans = l * r;
        for (i = 1;i <= n;i++)
        {
            if (a[i].x >= 0)
            {
                l = l + a[i].val;
                r = r - a[i].val;
            }
            else
            {
                l = l - a[i].val;
                r = r + a[i].val;
            }
            if (l * r > ans) ans = l * r;
        }
        printf("%lld\n",ans);
    } 
}

1010 Just do it

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

There is a nonnegative integer sequence

Input

The first line contains a positive integer

Output

For each test case:
A single line contains

Sample Input

1 1

3 3

1 2 3

Sample Output

1 3 1

题意：有一个长度为

题解：除了官方题解的卢卡斯定理，我们还可以发现每四次变换都是有规律的，因此每四次对其进行操作

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int t,n,m,i,k;
    int a[200050];
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        for (i = 1;i <= n;i++) scanf("%d",&a[i]);
        k = 1;
        while (k * 4 <= m) k = k * 4;
        while (m > 0)
        {
            while (m >= k)
            {
                for (i = k + 1;i <= n;i++)
                    a[i] = a[i] ^ a[i - k];
                m = m - k;
            }
            k = k / 4;
        }
        for (i = 1;i <= n;i++)
        {
            printf("%d",a[i]);
            if (i != n) printf(" "); else printf("\n");
        }
    }
}