标签：style blog color os io for ar 2014 log
题意：
给出k,x,n,m
找出这样的字符串s1,s2,s1长为n,s2长为m
给出规则，sn=sn-1+sn-2
使得第Sk项出现x个"AC"子串
我们很容易知道k次运算后有几个12，几个21，几个22子串，有几个串1，有几个串2
如果我们知道s1中出现了几个AC，s2中出现了几个AC
s1头和尾，s2头和尾巴
我们就能算出k次后能出现几个AC
现在正好反过来，我们都不知道s1,s2，我们要找出这样的s1,s2
那么我们枚举
s1头和尾，s2头和尾巴
s1中出现了几个AC，s2中出现了几个AC
算出来的Sk里面出现的AC数和x是否相等
直到枚举出来这样的
没有的话就输出happy new year
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
long long k,x,n,m;
long long lastnum[4][4];
long long num[4][4];
long long x1=1,x2=0;
long long xx1=0,xx2=1;
long long tmp[4][4];
void copy(long long to[][4],long long from[][4]){
for(long long i=1;i<=2;i++)
for(long long j=1;j<=2;j++)
to[i][j]=from[i][j];
}
void add(long long to[][4],long long from[][4]){
for(long long i=1;i<=2;i++)
for(long long j=1;j<=2;j++)
to[i][j]+=from[i][j];
}
long long ac_num(char* s){
long long cnt=0;
long long len=strlen(s);
for(long long i=0;i<len;i++)
if(s[i]=='A' && s[i]=='C') cnt++;
return cnt;
}
bool can(long long num1,long long num2,long long suf1,long long suf2,long long pre1,long long pre2){
if(xx1*num1+xx2*num2+
num[1][2]*(suf1=='A' && pre2=='C')+
num[2][2]*(suf2=='A' && pre2=='C')+
num[2][1]*(suf2=='A' && pre1=='C')
==x){
///cout<<"n::"<<num[1][2]<<' '<<num[2][2]<<' '<<num[2][1]<<endl;
return true;
}else
return false;
}
bool print(long long num1,long long num2, char suf1,char suf2,char pre1,char pre2){
if(n==1 && pre1 != suf1) return false;
if(m==1 && pre2 != suf2) return false;
if(n%2==0){ if(num1==n/2) if(!(suf1=='C' && pre1=='A')) return false;
}else if(num1==n/2)if ((!(pre1=='A'||suf1=='C'))&&(n!=1))return false;
if(m%2==0){ if(num2==m/2) if(!(suf2=='C' && pre2=='A')) return false;
}else if(num2==m/2) if ((!(pre2=='A'||suf2=='C'))&&(m!=1)) return false;
if(n%2==0){
if(num1==n/2){
for(long long i=1;i<=num1;i++) cout<<"AC";
}else{
cout<<pre1;
for(long long i=1;i<=num1;i++) cout<<"AC";
for(long long i=1;i<=n-num1*2-2;i++) cout<<"B";
cout<<suf1;
}
}else{
if(num1==n/2){
if(n==1) cout<<pre1;
else
if(pre1=='A'){
for(long long i=1;i<=num1;i++) cout<<"AC";
cout<<suf1;
}else if(suf1=='C'){
cout<<pre1;
for(long long i=1;i<=num1;i++) cout<<"AC";
}
}else{
cout<<pre1;
for(long long i=1;i<=num1;i++) cout<<"AC";
for(long long i=1;i<=n-num1*2-2;i++) cout<<"B";
cout<<suf1;
}
}
cout<<endl;
if(m%2==0){
if(num2==m/2){
for(long long i=1;i<=num2;i++) cout<<"AC";
}else{
cout<<pre2;
for(long long i=1;i<=num2;i++) cout<<"AC";
for(long long i=1;i<=m-num2*2-2;i++) cout<<"B";
cout<<suf2;
}
}else{
if(num2==m/2){
if(m==1) cout<<pre2;
else
if(pre2=='A'){
for(long long i=1;i<=num2;i++) cout<<"AC";
cout<<suf2;
}else if(suf2=='C'){
cout<<pre2;
for(long long i=1;i<=num2;i++) cout<<"AC";
}
}else{
cout<<pre2;
for(long long i=1;i<=num2-1;i++) cout<<"AC";
for(long long i=1;i<=m-num2*2-2;i++) cout<<"B";
cout<<suf2;
}
}
cout<<endl;
return true;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("G:/in.txt","r",stdin);
#endif
cin>>k>>x>>n>>m;
for(long long i=3;i<=k;i++){
long long xx1tmp=xx1,xx2tmp=xx2;
xx1+=x1;xx2+=x2;
x1=xx1tmp,x2=xx2tmp;
copy(tmp,num);
add(num,lastnum);
copy(lastnum,tmp);
if(i==3) num[1][2]=1;
if(i>=4 && i%2==0)
num[2][1]++;
else if(i>=4)
num[2][2]++;
}
for(long long num1=0;num1<=(n/2);num1++)
for(long long num2=0;num2<=(m/2);num2++)
for(long long suf1='A';suf1<='C';suf1++)
for(long long suf2='A';suf2<='C';suf2++)
for(long long pre1='A';pre1<='C';pre1++)
for(long long pre2='A';pre2<='C';pre2++)
if(can(num1,num2,suf1,suf2,pre1,pre2)){
///cout<<xx1<<' '<<xx2<<endl;
// cout<<num1<<' '
// <<num2<<' '
// <<(char)pre1<<' '
// <<(char)pre2<<' '
// <<(char)suf1<<' '
// <<(char)suf2<<' '<<endl;
if(print(num1,num2,suf1,suf2,pre1,pre2)){
return 0;
}
}
cout<<"Happy new year!"<<endl;
}
标签：style blog color os io for ar 2014 log
原文地址：http://blog.csdn.net/u011775691/article/details/38784041
