1 second
256 megabytes
standard input
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.
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 to 106 digits. The second line contains a pair of space-separated positive integers
a, b (1 ≤ a, b ≤ 108).
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).
116401024 97 1024
YES 11640 1024
284254589153928171911281811000 1009 1000
YES 2842545891539 28171911281811000
120 12 1
题目大意:把一个很大的数分成两部分,前一部分可以被a整除,后一部分可以被b整除,如果存在输出这两部分,两部分都不能含有前导0;
思路:从左到右求出在每一位对a的余数记录在aa数组里面,再从右向左在每一位对b的余数记录在bb数组里面。在aa[i]==0&&bb[i+1]==0&&s[i+1]!='0'的情况下,就存在答案。
代码:
# include<cstdio> # include<cstring> # include<algorithm> using namespace std; char str[2000000]; int a,b; int aa[2000000],bb[2000000]; int main(void) { while(scanf("%s",str)!=EOF) { scanf("%d%d",&a,&b); int k=strlen(str); int t=0; for(int i=0; i<k; i++) { t+=(str[i]-'0'); aa[i]=t%a; t%=a; t*=10; } int s=1,t1=0; for(int i=k-1; i>=0; i--) { bb[i]=((str[i]-'0')*s+t1)%b; t1=((str[i]-'0')*s+t1)%b; s=(s*10)%b; } bool flag=false; int pos=-1; for(int i=0; i<k-1; i++) { if(aa[i]==0&&bb[i+1]==0&&str[i+1]!='0') { pos=i; flag=true; break; } } if(flag) { printf("YES\n"); for(int i=0; i<=pos; i++) { printf("%c",str[i]); } printf("\n"); for(int i=pos+1; i<k; i++) { printf("%c",str[i]); } printf("\n"); } else { printf("NO\n"); } } return 0; }