题目描述：
n属于1到200，找到对应的一个数只含有0和1，并且是n的倍数；
分析：
本题有几个数会是大数；所以要考虑大数；
用到余数的性质；例如n=6，1%6=1；
1*10%6=4； （1*10+1）%6=5；
4*10%6=4； （4*10+1）%6=5；
5*10%6=2； （5*10+1）%6=3；
（重复4，5）
2*10%6=2； 。。。。=3；
3*10%6=0；
这时候发现余数为0，说明这个数可以是6的倍数；倒退回去，数分别是1，10，11，100，101，110，111，。。。。1110；
可以发现余数是一样的，同余定理；
（a*b）%n = （a%n *b%n）%n
（a+b）%n = （a%n +b%n）%n
由同余模定理 (110*10+1)%6 = (（110*10）%6+1%6 )%6 = (（110%6 * 10%6）%6 +1 )%6；
用这个同余定理就可以解决大数问题了；然后就是记录路径，这里就是巧妙的地方；我还不太清除是怎么搞的，总之就是一共进行了k次操作，就相当于01全排列，首项是1，然后排到第一个符合的数的时候，这个数是第几个，它对应的二进制就是相应的串；这一题用bfs居然超时了；所以我打了个表，有一个不打表的做法；
代码如下：
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 #include<queue>
6 using namespace std;
7
8 queue<int> q;
9 int n;
10 char a[205][200]={"1","10","111","100","10","1110","1001","1000","111111111","10","11","11100","1001","10010","1110","10000","11101","1111111110","11001","100","10101","110","110101","111000","100","10010","1101111111","100100","1101101","1110","111011","100000","111111","111010","10010","11111111100","111","110010","10101","1000","11111","101010","1101101","1100","1111111110","1101010","10011","1110000","1100001","100","100011","100100","100011","11011111110","110","1001000","11001","11011010","11011111","11100","100101","1110110","1111011111","1000000","10010","1111110","1101011","1110100","10000101","10010","10011","111111111000","10001","1110","11100","1100100","1001","101010","10010011","10000","1111111101","111110","101011","1010100","111010","11011010","11010111","11000","11010101","1111111110","1001","11010100","10000011","100110","110010","11100000","11100001","11000010","111111111111111111","100","101","1000110","11100001","1001000","101010","1000110","100010011","110111111100","1001010111","110","111","10010000","1011011","110010","1101010","110110100","10101111111","110111110","100111011","111000","11011","1001010","10001100111","11101100","1000","11110111110","11010011","10000000","100100001","10010","101001","11111100","11101111","11010110","11011111110","11101000","10001","100001010","110110101","100100","10011","100110","1001","1111111110000","11011010","100010","1100001","11100","110111","11100","1110001","11001000","10111110111","10010","1110110","1010100","10101101011","100100110","100011","100000","11101111","11111111010","1010111","1111100","1111110","1010110","11111011","10101000","10111101","111010","1111011111","110110100","1011001101","110101110","100100","110000","100101111","110101010","11010111","11111111100","1001111","10010","100101","110101000","1110","100000110","1001011","1001100","1010111010111","110010","11101111","111000000","11001","111000010","101010","110000100","1101000101","1111111111111111110","111000011","1000"};
11 /*
12 int cou=0;
13 void bfs()
14 {
15 while(!q.empty())
16 {
17 int t=q.front();
18 q.pop();
19 cou++;
20 if(t%n==0)
21 return;
22 q.push(t*10%n);
23 q.push((t*10+1)%n);
24 }
25 }
26 int main()
27 {
28 freopen("out","w",stdout);
29 for(int z=1;z<=200;z++)
30 {
31 n=z;
32 while(!q.empty())
33 q.pop();
34 memset(a,0,sizeof(a));
35 cou=0;
36 a[0]=1;
37 q.push(1);
38 bfs();
39 int i=0;
40 while(cou)
41 {
42 a[i++]=cou%2;
43 cou=cou/2;
44 }
45 printf("\"");
46 for(int j=i-1;j>=0;j--)
47 {
48 printf("%d",a[j]);
49 }
50 printf("\",");
51 }
52 return 0;
53 }*/
54
55 int main()
56 {
57 while(cin>>n&&n)
58 {
59 printf("%s\n",a[n-1]);
60 }
61 return 0;
62 }
1 //Memory Time
2 //2236K 32MS
3
4 #include<iostream>
5 using namespace std;
6
7 int mod[524286]; //保存每次mod n的余数
8 //由于198的余数序列是最长的
9 //经过反复二分验证，436905是能存储198余数序列的最少空间
10 //但POJ肯定又越界测试了...524286是AC的最低下限，不然铁定RE
11
12 int main(int i)
13 {
14 int n;
15 while(cin>>n)
16 {
17 if(!n)
18 break;
19
20 mod[1]=1%n; //初始化，n倍数的最高位必是1
21
22 for(i=2;mod[i-1]!=0;i++) //利用同余模定理，从前一步的余数mod[i/2]得到下一步的余数mod[i]
23 mod[i]=(mod[i/2]*10+i%2)%n;
24 //mod[i/2]*10+i%2模拟了BFS的双入口搜索
25 //当i为偶数时，+0，即取当前位数字为0 。为奇数时，则+1,即取当前位数字为1
26
27 i--;
28 int pm=0;
29 while(i)
30 {
31 mod[pm++]=i%2; //把*10操作转化为%2操作，逆向求倍数的每一位数字
32 i/=2;
33 }
34 while(pm)
35 cout<<mod[--pm]; //倒序输出
36 cout<<endl;
37 }
38 return 0;
39 }
时间： 2024-10-10 23:25:10