UVa:254 Towers of Hanoi

考试前最后一次码题了。

主要是利用移动n个盘子需要2^n-1步这个信息来简化整个过程。这恰好也是奇数步。如果n是奇数那么这些盘子都移动到当前的下一个柱子上，否则移动到隔一个的柱子上。

另外，第奇数步就是移动到下一个柱子上。第偶数步就是不移动1号盘子的时候，唯一只有一种可移动办法。

这样模拟整个过程就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ll long long
#define MAXN 50
using namespace std;
void BigSub(char *a,char*b,char*back)
{
    char c[MAXN];
    int down=0,p=0;
    for(int i=strlen(a)-1,j=strlen(b)-1; i>=0; --i,--j)
    {
        int x,y,z;
        x=a[i]-'0';
        if(j<0) y=0;
        else y=b[j]-'0';
        z=x-y+down;
        if(z<0)
        {
            down=-1;
            z+=10;
        }
        else down=0;
        c[p++]=z+'0';
    }
    c[p]=0;
    int q=0;
    bool ok=false;
    for(int i=p-1; i>=0; --i)
    {
        if(c[i]!='0') ok=true;
        if(ok) back[q++]=c[i];
    }
    if(!ok) back[q++]='0';
    back[q]=0;
}
char num[105][50]= {"0","1","3","7","15","31","63","127","255","511","1023","2047","4095","8191","16383","32767","65535","131071","262143","524287","1048575","2097151","4194303","8388607","16777215","33554431","67108863","134217727","268435455","536870911","1073741823","2147483647","4294967295","8589934591","17179869183","34359738367","68719476735","137438953471","274877906943","549755813887","1099511627775","2199023255551","4398046511103","8796093022207","17592186044415","35184372088831","70368744177663","140737488355327","281474976710655","562949953421311","1125899906842623","2251799813685247","4503599627370495","9007199254740991","18014398509481983","36028797018963967","72057594037927935","144115188075855871","288230376151711743","576460752303423487","1152921504606846975","2305843009213693951","4611686018427387903","9223372036854775807","18446744073709551615","36893488147419103231","73786976294838206463","147573952589676412927","295147905179352825855","590295810358705651711","1180591620717411303423","2361183241434822606847","4722366482869645213695","9444732965739290427391","18889465931478580854783","37778931862957161709567","75557863725914323419135","151115727451828646838271","302231454903657293676543","604462909807314587353087","1208925819614629174706175","2417851639229258349412351","4835703278458516698824703","9671406556917033397649407","19342813113834066795298815","38685626227668133590597631","77371252455336267181195263","154742504910672534362390527","309485009821345068724781055","618970019642690137449562111","1237940039285380274899124223","2475880078570760549798248447","4951760157141521099596496895","9903520314283042199192993791","19807040628566084398385987583","39614081257132168796771975167","79228162514264337593543950335","158456325028528675187087900671","316912650057057350374175801343","633825300114114700748351602687","1267650600228229401496703205375"};
int Compare(char* a,char* b) //a>b 返回1,a<b返回-1,相同返回0
{
    int la=strlen(a),lb=strlen(b);
    if(la>lb) return 1;
    else if(la<lb) return -1;
    else  return strcmp(a,b);
}
int Search(int low,int high,char *key)
{
    int mid=(low+high+1)/2;
    while(low<high)
    {
        if(Compare(num[mid],key)<=0) low=mid;
        else high=mid-1;
        mid=(low+high+1)/2;
    }
    return mid;
}
int v[5];
char cnt[MAXN];
void Move(int a)
{

    int s=Search(0,100,cnt);
    char step[MAXN];
    strcpy(step,num[s]);
    int x,y;
    if(s%2)
    {
        x=1;
        y=2;
    }
    else
    {
        x=2;
        y=1;
    }
    v[a%3]-=s;
    v[(a+x)%3]+=s;
    BigSub(cnt,step,cnt);
    if(!strcmp(cnt,"0")) return ;
    else
    {
        v[a%3]--;
        v[(a+y)%3]++;
        BigSub(cnt,"1",cnt);
        if(!strcmp(cnt,"0")) return ;
        else Move((a+x)%3);
    }
}
int main()
{
    //freopen("outs.txt","w",stdout);
    int n;
    while(scanf("%d%s",&n,cnt))
    {
        if(!n&&!strcmp(cnt,"0")) break;
        v[0]=v[1]=v[2]=0;
        v[0]=n;
        Move(0);
        printf("%d %d %d\n",v[0],v[1],v[2]);
    }
    return 0;
}