[Problem of] the original title Portal
[Effect] problem solution
This problem solution written in a very clear and easy to understand, I will not go into details.
Talk about the thing I read in the above blog is not written out directly but may affect the understanding of things:
f(x) = f(x/5) × a(x % 20)
Known non-zero after the last one must be an even number, the number 20 we consider each into a group, so that they are multiplied == 4 * 4% 10 6
(No matter how much is, no effect on the product are not behind a number of points.
Algorithm to achieve:
We give the number denoted by x, since the answer to ans [x].
1% high-precision single-precision (20) = y;
2. record the answers as ans [x] = a [y] * ans [x / 5];
years [x] = a [y] * years [x / 5] = a [y 1] * a [y 2] * years [x / 5/5] = ...
Until x <20.
About a pre-array should be:
4, 1 / 2,5,6,7,8,9,1 / 2,11,12,13,14,1 / 2,16,17,18,19,1 / 2 take the time to remember the prefix and 10%.
Each packet number 20, the number less than the number of the sub-group number less than 20. That number is less than 20 we require pretreatment.
【code】
#include<bits/stdc++.h> using namespace std; #define File "" #define ll long long inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } const int mxn = 110; char s[mxn]; int T,a[mxn],b[22],n; inline int calc1(){ return (a[2]*10+a[1]) % 20; } inline void calc2(){ int t = 0; for(int i = n;i >= 1; --i){ t = t*10+a[i]; a[i] = t/5; t = t%5; } } int ans; int main(){ // file(); T = 5,b[0] = 1; for(int i = 1;i <= 20; ++i){ if(!(i%5)) b[i] = b[i-1]/2%10; else b[i] = b[i-1]*i%10; } // for(int i = 1;i <= 20; ++i) printf("%d ",b[i]); // puts(""); while(T--){ scanf("%s",s+1); n = strlen(s+1); for(int i = 1;i <= n; ++i) a[n-i+1] = s[i]-'0'; ans = 1; while(n > 0){ while(a[n]==0) n--; year = years * b [calc1 ()]% 10 ; calc2(); } printf("%d\n",ans); } return 0; } /* 11 12 13 14 15 */