2012 Multi-University Training Contest 6

官方解题报告http://page.renren.com/601081183/note/865145486?null&ref=minifeed&sfet=2011&fin=0&ff_id=601081183&feed=page_reblog&tagid=1981432999&statID=page_601081183_2&level=2

 

1001  hdu 4350Card

 http://acm.hdu.edu.cn/showproblem.php?pid=4350

题意：

给定52张牌，按顺序排列着，给出操作每次讲下表为[l,r]的牌一道最前边，问执行n次这样的操作后最终序列的情况。

思路：

给定的n比较大，直接模拟肯定tle,首先我们发现移动m次[l,r]后肯定会出现循环的情况，所以我们只要找出循环节，然后n %= m 然后剩下的模拟就可以；

View Code

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <vector>

#define maxn 66

using namespace std;



int a[maxn],b[maxn];



int gcd(int a,int b)

{

    if (b == 0) return a;

    else return gcd(b,a%b);

}

int main()

{

    //freopen("d.txt","r",stdin);

    int L,R,i,j;

    int t,n;

    int cas = 1;

    scanf("%d",&t);

    while (t--)

    {

        for (i = 1; i <= 52; ++i) scanf("%d",&a[i]);

        //for (i = 1; i <= 52; ++i) printf("%d ",a[i]);

        scanf("%d%d%d",&n,&L,&R);

        int d = ((R - L + 1)*R)/gcd(R - L + 1,R);//找出循环节

        n %= d;

        while (n--)//模拟

        {



           j = 1;

           int len = R - L + 1;

           for (i = L; i <= R; ++i) b[j++] = a[i];

           for (i = L - 1; i >= 1; --i) a[i + len] = a[i];

           for (i = 1; i < j; ++i)

           a[i] = b[i];

        }

        printf("Case #%d:",cas++);

        for (i = 1; i <= 52; ++i) printf(" %d",a[i]);

        printf("\n");

    }

    return 0;

}

 

1006 hdu 4355 Party All the Time

http://acm.hdu.edu.cn/showproblem.php?pid=4355

题意：

在x坐标轴上给出n个点的x坐标以及他们本身的权值w,求能够找出一点Xm是的所有点到该店的距离Si Si^3*wi的和最小。

思路：

才开始我以为找到中点，前后枚举100个点蹭一下数据可以过，可是一看这里x坐标是小数啊，你怎么枚举，只好错误的做了一下结果真的wa.这题写出公式函数是一个凹函数，直接套三分模板即可。

View Code

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <vector>

#include <cmath>

#define maxn 50007

using namespace std;



const double inf = 0x7fffffff;

const double eps = 1e-4;



struct node

{

    double x,w;

}p[maxn];

int n;



int dbcmp(double x)

{

    if (x > eps) return 1;

    else if (x < -eps) return -1;

    else return 0;

}

double getS(int x,double y)

{

    return fabs((p[x].x - y)*(p[x].x - y)*(p[x].x - y))*p[x].w;

}

double getR(double x)

{

    double sum = 0;

    for (int i = 0; i < n; ++i)

    {

        sum += getS(i,x);

    }

    return sum;

}

void solve(double l,double r,int cas)

{

    double m,mm;

    while (dbcmp(r - l) > 0)

    {

        m = (l + r)/2.0;

        mm = (m + r)/2.0;

        if (getR(m) >= getR(mm)) l = m;

        else r = mm;

    }

    printf("Case #%d: %.0lf\n",cas,getR(l));

}

int main()

{

    //freopen("d.txt","r",stdin);

    int t,i;

    int cas = 1;

    scanf("%d",&t);

    while (t--)

    {

        scanf("%d",&n);

        double l = inf;

        double r = -inf;

        for (i = 0; i < n; ++i)

        {

            scanf("%lf%lf",&p[i].x,&p[i].w);

            l = min(l,p[i].x);

            r = max(r,p[i].x);

        }

        solve(l,r,cas);

        cas++;

    }

    return 0;

}

 

1008 hdu4357String change

 http://acm.hdu.edu.cn/showproblem.php?pid=4357

 

题意：

给你两个字符串，s1,s2,问是都能够通过交换s1里面的任意两个字符串并且同时字符串要+1，得到s2;

思路：

官方解题报告有证明不在多说，给代码，MB在最后的几分钟写的，结果把两数交换给写错了，导致最后没有AC。。悲剧啊。。

View Code

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <vector>

#define maxn 66

using namespace std;



char s1[maxn],s2[maxn];



int main()

{

    //freopen("d.txt","r",stdin);

    int t;

    int i;

    int cas = 1 ;

    scanf("%d",&t);

    while (t--)

    {

        scanf("%s%s",s1,s2);

        int len = strlen(s1);

        if (len == 2)

        {

            int a1 = s1[0] - 'a' + 1;

            int b1 = s1[1] - 'a' + 1;

            int a2 = s2[0] - 'a' + 1;

            int b2 = s2[1] - 'a' + 1;

            i = 1;

            while (i <= 26)

            {

                if (a1 == a2 && b1 == b2) break;

                a1++; b1++;

                if (a1 > 26) a1 %= 26;

                if (b1 > 26) b1 %= 26;

                int tmp = a1;

                a1 = b1;

                b1 = tmp;

                i++;

            }

            if (a1 == a2 && b1 == b2) printf("Case #%d: YES\n",cas++);

            else  printf("Case #%d: NO\n",cas++);



        }

        else

        {

            int sum1 = 0,sum2 = 0;

            for (i = 0; i < len; ++i)

            {

                sum1 += s1[i] - 'a' + 1;

                sum2 += s2[i] - 'a' + 1;

            }

            if ((sum1%2 == 0 && sum2%2 == 0) || (sum1%2 == 1 && sum2%2 == 1))

            {

                printf("Case #%d: YES\n",cas++);

            }

            else

            {

                printf("Case #%d: NO\n",cas++);

            }

        }

    }

    return 0;

}