Codeforces Round #279 (Div. 2) C. Hacking Cypher （大数取余）

题目链接

C. Hacking Cypher
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.

Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!

Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by a as a separate number, and the second (right) part is divisible by b as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values a and b.

Help Polycarpus and find any suitable method to cut the public key.

Input
The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to106 digits. The second line contains a pair of space-separated positive integers a, b (1 ≤ a, b ≤ 108).

Output
In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by a, and the right part must be divisible by b. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.

If there is no answer, print in a single line "NO" (without the quotes).

Sample test(s)
input
116401024
97 1024
output
YES
11640
1024
input
284254589153928171911281811000
1009 1000
output
YES
2842545891539
28171911281811000
input
120
12 1
output
NO

 

题意：一个很大的数，可能有10^6位，给两个数字a,b 求把给的大数分为两部分，如果这两部分
能分别整除a,b，且没有前导0，输出YES,否则输出NO.

分析：大数取余。

正着一遍a, 倒着一遍b。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cmath>
 6 #include <map>
 7 #include <vector>
 8 #include <algorithm>
 9 #define LL __int64
10 const int maxn = 1e6 + 10;
11 using namespace std;
12 char s[maxn];
13 int f1[maxn], f2[maxn];
14 
15 int main()
16 {
17     int a, b, i, len;
18     int tmp, tmp2, x;
19     int y;
20     while(~scanf("%s", s))
21     {
22         scanf("%d%d", &a, &b);
23         memset(f1, 0, sizeof(f1));
24         memset(f2, 0, sizeof(f2));
25         len = strlen(s);
26         tmp2 = 0;
27         for(i = 0; i < len; i++)
28         {
29             tmp = (s[i]-48)%a;
30             tmp2 = ((tmp2*10)+tmp)%a;
31             if(tmp2==0) f1[i] = 1;
32         }
33 
34         tmp2 = 0; x = 1;
35         for(i = len-1; i >= 0; i--)
36         {
37             tmp = (s[i]-48)%b;
38             tmp = (tmp*x)%b;
39             tmp2 = (tmp2 + tmp)%b;
40             if(tmp2 == 0) f2[i] = 1;
41             x = (x*10)%b;
42         }
43 
44         y = -1;
45         for(i = 0; i < len-1; i++)
46         {
47             if(f1[i]&&f2[i+1] && s[i+1]!='0')
48             {
49                 y = i;
50                 break;
51             }
52         }
53         if(y==-1)
54         cout<<"NO"<<endl;
55         else
56         {
57             cout<<"YES"<<endl;
58             for(i = 0; i <= y; i++)
59             printf("%c", s[i]);
60             cout<<endl;
61             for(; i < len; i++)
62             printf("%c", s[i]);
63             cout<<endl;
64         }
65     }
66     return 0;
67 }

 

转载于:https://www.cnblogs.com/bfshm/p/4119519.html