TOJ 1086 Round and Round We Go 大数相乘

亦凉 2022-05-08 07:48 86阅读 0赞

## **1086: Round and Round We Go** ##

**时间限制(普通/Java):1000MS/10000MS 内存限制:65536KByte  
总提交: 254 测试通过:165**

**描述**

A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a"cycle"of the digits of the original number. That is, if you consider the number after the last digit to "wrap around"back to the first digit, the sequence of digits in both numbers will be the same, though they may start at different positions.For example, the number 142857 is cyclic, as illustrated by the following table:  
142857 \*1 = 142857  
142857 \*2 = 285714  
142857 \*3 = 428571  
142857 \*4 = 571428  
142857 \*5 = 714285  
142857 \*6 = 857142

**输入**

Write a program which will determine whether or not numbers are cyclic. The input file is a list of integers from 2 to 60 digits in length. (Note that preceding zeros should not be removed, they are considered part of the number and count in determining n. Thus, "01"is a two-digit number, distinct from "1" which is a one-digit number.)

**输出**

For each input integer, write a line in the output indicating whether or not it is cyclic.

**样例输入**

142857  
142856  
142858  
01  
0588235294117647

**样例输出**

142857 is cyclic  
142856 is not cyclic  
142858 is not cyclic  
01 is not cyclic  
0588235294117647 is cyclic

**题目来源**

[Greater New York 200][]

大数向量相乘，上半部分代码是Karatsuba算法 的板子，下面是对这题的判断，循环数最多只要\*9，因为\*10的话位数肯定会增加就不是循环数啦。

#include<iostream>
    using namespace std;
    #include<vector>
    #include<math.h>
    vector<int> a,b;
    // 向量相加 
    vector<int> Add(vector<int> &a,vector<int> &b,int k)
    {
        for(int i=0;i<k;i++) b.push_back(0);    //b*10^k
        int asize = a.size();
        int bsize = b.size();
        if(asize<bsize) return Add(b,a,0);  //a比b短   
        int flag,n,diff;
        flag = 0;   //进位 
        diff = asize - bsize; //相差位数 
        for(int i=asize-1;i>=0;i--){    //从最后一位开始加 
            if(i-diff >= 0)n = a[i]+b[i-diff]+flag;
            else n=a[i]+flag;
            flag = n/10;
            a[i] = n%10;
    
        }
        if(flag > 0){   //处理最高位进位 
            a.push_back(1);
            for(int i=asize;i>=1;i--){
                a[i] = a[i-1];
            }
            a[0] = flag;
        }
        return a;
    }
    // 向量相减 
    vector<int> Sub(vector<int> &a,vector<int> &b)
    {
        int asize = a.size();
        int bsize = b.size();
        if(asize<bsize) return Sub(b,a);    //a比b短 
        int flag,n,diff;
        flag = 0;
        diff = asize - bsize;   //相差位数 
        for(int i=asize-1;i>=0;i--){
            if(i-diff >= 0)n = a[i]-b[i-diff]+flag;
            else n=a[i]+flag;
            if(a[i]-b[i-diff] < 0 && i-diff >=0){
                flag = -1;
                a[i] = 10+n;
            }
            else{
                flag = 0;
                a[i] = n;
            }
        }
        int cntzero = 0;    //高位的零的个数 
        for(int i=0;i<a.size();i++){
            if(a[i] == 0) cntzero++;
            else break;
        }
        for(int i=0;i<a.size();i++) //消除高位的零 
            a[i] = a[i+cntzero];
        for(int i=0;i<cntzero;i++)//删除多余部分 
            a.pop_back();
    
        return a;
    }
    
    //向量乘法
     vector<int> Multi(vector<int> &a,vector<int> &b)
    {
        if(a.size()>b.size()) 
    		return Multi(b,a);    //a比b大 
        vector<int> v(a.size() + b.size() +1);  //结果数组 v 
        int diff = b.size()-a.size();
        for(int i=0;i<a.size();i++)
    	{   //循环相加 
            for(int j=0;j<b.size();j++)
            v[i+j] += a[a.size()-1-i] * b[b.size()-1-j];
         }  
    	for(int i=0;i<v.size();i++)
    	{
             if(v[i]<0)
    		 {
                int borrow = (abs(v[i]) + 9) /10;
                v[i+1] -= borrow;
                v[i] +=borrow*10;
             }
             else
     		 {
                v[i+1] += v[i] /10;
                v[i] %= 10;
             }  
         }
         while(v.size() > 1&&v.back() == 0) 
    	 	v.pop_back();//去除 v 多于部分 
         vector<int> c(v.size());
         for(int i=0;i<v.size();i++)
    	 {
            c[i] = v[v.size()-1 - i];   //倒序输出 
         }
         return c;
     }
    // Karatsuba算法 
    vector<int> Karatsuba(vector<int> &a,vector<int> &b)
    {
        int asize = a.size();
        int bsize = b.size();
        if(asize <2 || bsize <2) 
    		return Multi(a,b); //长度小于2 使用普通乘法 
        if(asize < bsize) 
    		return Karatsuba(b,a);    //保证 a长度比b大 
        if(asize == 0 || bsize == 0) 
    		return vector<int>();
        int half = asize / 2;   //去一半 
        // a 分成 a1(高位),a0(低位) 
        vector<int> a1(a.begin(),a.end()-half);
        vector<int> a0(a.end()-half,a.end());
        //同理 b 
        vector<int> b1(b.begin(),b.end()-min(half,bsize));
        vector<int> b0(b.end()-min(half,bsize),b.end());
    
        vector<int> z0 = Karatsuba(a0,b0);
        vector<int> z2 = Karatsuba(a1,b1);
        vector<int> tmpa = Add(a0,a1,0),tmpb = Add(b0,b1,0);
        vector<int> z1 = Karatsuba(tmpa,tmpb);          
    
        z1 = Sub(z1,z0);
        z1 = Sub(z1,z2);
    
        vector<int> ret;
        ret.push_back(0);
    
        ret = Add(ret,z0,0);
        ret = Add(ret,z1,half);
        ret = Add(ret,z2,half*2);
    
        return ret;
    }
    int main()
    {
        char x[100];
        int i,j,l,p,len,flag;
        while(scanf("%s",x)!=EOF)
    	{
    		a.clear();
    		b.clear();
    		flag=0;
    		len=strlen(x);
    		for(i=0;i<len;i++)
    		a.push_back(x[i]-'0');
    		for(i=2;i<=len;i++)
    		{
    			if(i>=10)break;
    			b.clear();
    			if(i<10)
    			b.push_back(i);
    			else
    			{
    				b.push_back(i/10);
    				b.push_back(i%10);
    			}			
    			/*for(j=0;j<b.size();j++)
    			printf("%d ",b[j]);
    			printf("***\n");*/
    			vector<int> v=Karatsuba(a,b);
    			int n=v.size();
    			/*for(j=0;j<n;j++)
    			printf("%d",v[j]);
    			printf("\n****\n");*/
    			int k=0;
    			for(p=0;p<len;p++)
    			{
    				if((x[p]-'0')==v[0])
    				{
    					k=0;
    					for(j=p,l=0;l<len;j++,l++)
    					{
    						if((x[j%len]-'0')==v[l])k++;
    						else
    						{
    							k=0;
    							break;
    						}
    					}
    					if(k==len)break;
    				}
    			}
    			if(k==len)continue;
    			else
    			{
    				flag=1;
    				break;
    			}
    		}
    		if(flag)
    		printf("%s is not cyclic\n",x);
    		else
    		printf("%s is cyclic\n",x);
    	}
        
    }

奋斗完之后，发现学姐的代码的是这样的，姜的还是老的辣...

#include<stdio.h>
    #include<algorithm>
    int main()
    {
    	char a[62];
    	int i,n,s,t;
    	while(~scanf("%s",a))
    	{
    		n=strlen(a);
    		for(t=0,i=n-1;i>=0;i--)
    		{
    			s=(a[i]-'0')*(n+1)+t;
    			t=s/10;s=s%10;
    			if(s!=9)break;
    		}
    		printf("%s is%s cyclic\n",a,i+1?" not":"");
    	}
    }

[Greater New York 200]: http://210.32.82.1/acmhome/search.do?method=simpleSearch&searchFrom=Source&queryString=Greater+New+York+2001