BestCoder Round #29——A--GTY's math problem（快速幂（对数法））、B--GTY's birthday gift（矩阵快速幂）

GTY's math problem

Time Limit: 1000/1000 MS
(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

GTY is a GodBull who will get an Au in
NOI . To have more time to learn algorithm knowledge, he never does his math
homework. His math teacher is very unhappy for that, but she can't do anything
because GTY can always get a good mark in math exams. One day, the math teacher
asked GTY to answer a question. There are four numbers on the blackboard - a,b,c,d. The math teacher wants GTY to compare ab with cd. Because GTY never does his homework,
he can't figure out this problem! If GTY can't answer this question correctly,
he will have to do his homework. So help him!

Input

Multi test cases (about 5000). Every
case contains four integers a,b,c,d(1≤a,b,c,d≤1000)separated by spaces. Please process to
the end of file.

Output

For each case , if ab>cd , print '>'. if ab<cd , print '<'. if ab=cd , print '='.

Sample Input

2 1 1 2

2 4 4 2

10 10 9 11

Sample Output

>

=

<

GTY's math problem

Time Limit: 1000/1000 MS
(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1636    Accepted Submission(s): 298

问题描述

众所周知，GTY是一位神犇，为了更好的虐场，他从来不写数学作业而是去屠题，他的数学老师非常不爽，但由于GTY每次考试都AK，她也不能说什么，有一天老师在黑板上写了四个数——a,b,c,d 然后让GTY比较ab和cd的大小，由于GTY不屑于虐这道题，就把这个问题交给你了。

输入描述

多组数据（约5000组），每组数据包含4个整数a,b,c,d(1≤a,b,c,d≤1000)，用空格隔开

输出描述

对于每组数据，若ab>cd，输出”>”， 若ab<cd，输出”<”， 若ab=cd，输出”=”

输入样例

2 1 1 2

2 4 4 2

10 10 9 11

输出样例

>

=

<

思路  log(a^b)=b∗log(a)  注意精度！

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
#define emp (1e-8)
int main()
{
    double a,b,c,d;
    while(scanf("%lf%lf%lf%lf",&a,&b,&c,&d)!=EOF)
    {
        if(fabs(b*log(a)-d*log(c))<=emp)
            printf("=\n");
        else if(b*log(a)>d*log(c))
            printf(">\n");
        else if(b*log(a)<d*log(c))
            printf("<\n");
    }
    return ;
}

GTY's birthday gift

Time Limit: 2000/1000 MS
(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

FFZ's birthday is coming. GTY wants to
give a gift to ZZF. He asked his gay friends what he should give to ZZF. One of
them said, 'Nothing is more interesting than a number multiset.' So GTY decided
to make a multiset for ZZF. Multiset can contain elements with same values.
Because GTY wants to finish the gift as soon as possible, he will use JURUO
magic. It allows him to choose two numbers a and b(a,b∈S), and add a+b to the multiset. GTY can use the magic
for k times, and he wants the sum of the multiset is maximum, because the
larger the sum is, the happier FFZ will be. You need to help him calculate the
maximum sum of the multiset.

Input

Multi test cases (about 3) . The first
line contains two integers n and k (2≤n≤100000,1≤k≤1000000000). The second line contains n elements ai (1≤ai≤100000)separated by spaces , indicating the
multiset S .

Output

For each case , print the maximum sum of
the multiset (mod 10000007).

Sample Input

3 2

3 6 2

Sample Output

35

GTY's birthday gift

Time Limit: 2000/1000 MS
(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 585    Accepted Submission(s): 103

问题描述

GTY的朋友ZZF的生日要来了，GTY问他的基友送什么礼物比较好，他的一个基友说送一个可重集吧！于是GTY找到了一个可重集S，GTY能使用神犇魔法k次，每次可以向可重集中加入一个数 a+b(a,b∈S)，现在GTY想最大化可重集的和，这个工作就交给你了。

注：可重集是指可以包含多个相同元素的集合

输入描述

多组数据（约3组），每组数据的第一行有两个数n,k(2≤n≤100000,1≤k≤1000000000) 表示初始元素数量和可使用的魔法数，第二行包含n个数a(1≤ai≤100000)表示初始时可重集的元素

输出描述

对于每组数据，模10000007输出可重集可能的最大和。

输入样例

3 2

3 6 2

输出样例

35

这题是典型的矩阵快速幂：

这是公式：

#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL __int64
#define mod 10000007

int N;

struct matrix
{
    LL m[][];
};
int a[];

matrix multiply(matrix x,matrix y)
{
    matrix temp;
    memset(temp.m,,sizeof(temp.m));
    for(int i=; i<; i++)
    {
        for(int j=; j<; j++)
        {
            if(x.m[i][j]==) continue;
            for(int k=; k<; k++)
            {
                if(y.m[j][k]==) continue;
                temp.m[i][k]+=x.m[i][j]*y.m[j][k]%mod;
                temp.m[i][k]%=mod;
            }
        }
    }
    return temp;
}

matrix quickmod(matrix a,int n)
{
    matrix res;//单位阵
    memset(res.m,,sizeof(res.m));
    for(int i=;i<;i++) res.m[i][i]=;
    while(n)
    {
        if(n&)
            res=multiply(res,a);
        n>>=;
        a=multiply(a,a);
    }
/*    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
            printf("%8d",res.m[i][j]);
        printf("\n");
    }
    printf("\n");*/
    return res;
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        LL sum=;
        for(int i=;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n);
        matrix ans;
        ans.m[][]=;ans.m[][]=;ans.m[][]=;
        ans.m[][]=;ans.m[][]=;ans.m[][]=;
        ans.m[][]=;ans.m[][]=;ans.m[][]=;
        ans=quickmod(ans,k);

        printf("%d\n",(ans.m[][]*sum+ans.m[][]*a[n-]+ans.m[][]*a[n-])%mod);
    }
    return ;
}

今天忘记报名BC了，只能现在做了。才做了2题！还要加油啊。

矩阵快速幂，还得谢谢庆神当初的传授啊。。。。哈哈！