HDU 5351 MZL's Border(大数 + 找规律)——2015 Multi-University Training Contest 5
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MZL’s Border
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1374 Accepted Submission(s): 452
Problem Description
As is known to all, MZL is an extraordinarily lovely girl. One day, MZL was playing with her favorite data structure, strings.
MZL is really likeFibonacci Sequence , so she defines Fibonacci Strings in the similar way. The definition of Fibonacci Strings is given below.
1)fib1=b
2)fib2=a
3)fibi=fibi−1fibi−2, i>2
For instance,fib3=ab, fib4=aba, fib5=abaab .
Assume that a strings whose length is n is s1s2s3...sn . Then sisi+1si+2si+3...sj is called as a substring of s , which is written as s[i:j] .
Assume thati<n . If s[1:i]=s[n−i+1:n] , then s[1:i] is called as a Border of s . In Borders of s , the longest Border is called as s ' LBorder . Moreover, s[1:i] 's LBorder is called as LBorderi .<br><br>Now you are given 2 numbers n and m . MZL wonders what LBorderm of fibn is. For the number can be very big, you should just output the number modulo 258280327(=2×317+1) .<br><br>Note that 1≤T≤100, 1≤n≤103, 1≤m≤|fibn| .</div><div class=panel_bottom> </div><br><div class=panel_title align=left>Input</div> <div class=panel_content>The first line of the input is a number T , which means the number of test cases.<br><br>Then for the following T lines, each has two positive integers n and m , whose meanings are described in the description.</div><div class=panel_bottom> </div><br><div class=panel_title align=left>Output</div> <div class=panel_content>The output consists of T lines. Each has one number, meaning fibn 's LBorderm modulo 258280327(=2×317+1) .
MZL is really like
1)
2)
3)
For instance,
Assume that a string
Assume that
Sample Input
2
4 3
5 5
Sample Output
1
2
Author
SXYZ
Source
题目大意:
给出一个类似斐波那契数列的字符串序列:
输入一个
解题思路:
在纸上画一画,就可以发现规律是
其中
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; #define DIGIT 4 //四位隔开,即万进制 #define DEPTH 10000 //万进制 #define MAX 2000+5 //题目最大位数/4,要不大直接设为最大位数也行 typedef int bignum_t[MAX+1]; /************************************************************************/ /* 读取操作数,对操作数进行处理存储在数组里 */ /************************************************************************/ int read(bignum_t a,istream&is=cin) { char buf[MAX*DIGIT+1],ch ; int i,j ; memset((void*)a,0,sizeof(bignum_t)); if(!(is>>buf))return 0 ; for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--) ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ; for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0'); for(i=1;i<=a[0];i++) for(a[i]=0,j=0;j<DIGIT;j++) a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ; for(;!a[a[0]]&&a[0]>1;a[0]--); return 1 ; } void write(const bignum_t a,ostream&os=cout) { int i,j ; for(os<<a[i=a[0]],i--;i;i--) for(j=DEPTH/10;j;j/=10) os<<a[i]/j%10 ; } int comp(const bignum_t a,const bignum_t b) { int i ; if(a[0]!=b[0]) return a[0]-b[0]; for(i=a[0];i;i--) if(a[i]!=b[i]) return a[i]-b[i]; return 0 ; } int comp(const bignum_t a,const int b) { int c[12]= { 1 } ; for(c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++); return comp(a,c); } int comp(const bignum_t a,const int c,const int d,const bignum_t b) { int i,t=0,O=-DEPTH*2 ; if(b[0]-a[0]<d&&c) return 1 ; for(i=b[0];i>d;i--) { t=t*DEPTH+a[i-d]*c-b[i]; if(t>0)return 1 ; if(t<O)return 0 ; } for(i=d;i;i--) { t=t*DEPTH-b[i]; if(t>0)return 1 ; if(t<O)return 0 ; } return t>0 ; } /************************************************************************/ /* 大数与大数相加 */ /************************************************************************/ void add(bignum_t a,const bignum_t b) { int i ; for(i=1;i<=b[0];i++) if((a[i]+=b[i])>=DEPTH) a[i]-=DEPTH,a[i+1]++; if(b[0]>=a[0]) a[0]=b[0]; else for(;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++); a[0]+=(a[a[0]+1]>0); } /************************************************************************/ /* 大数与小数相加 */ /************************************************************************/ void add(bignum_t a,const int b) { int i=1 ; for(a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++); for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++); } /************************************************************************/ /* 大数相减(被减数>=减数) */ /************************************************************************/ void sub(bignum_t a,const bignum_t b) { int i ; for(i=1;i<=b[0];i++) if((a[i]-=b[i])<0) a[i+1]--,a[i]+=DEPTH ; for(;a[i]<0;a[i]+=DEPTH,i++,a[i]--); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数减去小数(被减数>=减数) */ /************************************************************************/ void sub(bignum_t a,const int b) { int i=1 ; for(a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++); for(;!a[a[0]]&&a[0]>1;a[0]--); } void sub(bignum_t a,const bignum_t b,const int c,const int d) { int i,O=b[0]+d ; for(i=1+d;i<=O;i++) if((a[i]-=b[i-d]*c)<0) a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ; for(;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数相乘,读入被乘数a,乘数b,结果保存在c[] */ /************************************************************************/ void mul(bignum_t c,const bignum_t a,const bignum_t b) { int i,j ; memset((void*)c,0,sizeof(bignum_t)); for(c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++) for(j=1;j<=b[0];j++) if((c[i+j-1]+=a[i]*b[j])>=DEPTH) c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ; for(c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--); } /************************************************************************/ /* 大数乘以小数,读入被乘数a,乘数b,结果保存在被乘数 */ /************************************************************************/ void mul(bignum_t a,const int b) { int i ; for(a[1]*=b,i=2;i<=a[0];i++) { a[i]*=b ; if(a[i-1]>=DEPTH) a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ; } for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++); for(;!a[a[0]]&&a[0]>1;a[0]--); } void mul(bignum_t b,const bignum_t a,const int c,const int d) { int i ; memset((void*)b,0,sizeof(bignum_t)); for(b[0]=a[0]+d,i=d+1;i<=b[0];i++) if((b[i]+=a[i-d]*c)>=DEPTH) b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ; for(;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH); for(;!b[b[0]]&&b[0]>1;b[0]--); } /**************************************************************************/ /* 大数相除,读入被除数a,除数b,结果保存在c[]数组 */ /* 需要comp()函数 */ /**************************************************************************/ void div(bignum_t c,bignum_t a,const bignum_t b) { int h,l,m,i ; memset((void*)c,0,sizeof(bignum_t)); c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ; for(i=c[0];i;sub(a,b,c[i]=m,i-1),i--) for(h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1) if(comp(b,m,i-1,a))h=m-1 ; else l=m ; for(;!c[c[0]]&&c[0]>1;c[0]--); c[0]=c[0]>1?c[0]:1 ; } void div(bignum_t a,const int b,int&c) { int i ; for(c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数平方根,读入大数a,结果保存在b[]数组里 */ /* 需要comp()函数 */ /************************************************************************/ void sqrt(bignum_t b,bignum_t a) { int h,l,m,i ; memset((void*)b,0,sizeof(bignum_t)); for(i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--) for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1) if(comp(b,m,i-1,a))h=m-1 ; else l=m ; for(;!b[b[0]]&&b[0]>1;b[0]--); for(i=1;i<=b[0];b[i++]>>=1); } /************************************************************************/ /* 返回大数的长度 */ /************************************************************************/ int length(const bignum_t a) { int t,ret ; for(ret=(a[0]-1)*DIGIT,t=a[a[0]];t;t/=10,ret++); return ret>0?ret:1 ; } /************************************************************************/ /* 返回指定位置的数字,从低位开始数到第b位,返回b位上的数 */ /************************************************************************/ int digit(const bignum_t a,const int b) { int i,ret ; for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--); return ret%10 ; } /************************************************************************/ /* 返回大数末尾0的个数 */ /************************************************************************/ int zeronum(const bignum_t a) { int ret,t ; for(ret=0;!a[ret+1];ret++); for(t=a[ret+1],ret*=DIGIT;!(t%10);t/=10,ret++); return ret ; } void comp(int*a,const int l,const int h,const int d) { int i,j,t ; for(i=l;i<=h;i++) for(t=i,j=2;t>1;j++) while(!(t%j)) a[j]+=d,t/=j ; } void convert(int*a,const int h,bignum_t b) { int i,j,t=1 ; memset(b,0,sizeof(bignum_t)); for(b[0]=b[1]=1,i=2;i<=h;i++) if(a[i]) for(j=a[i];j;t*=i,j--) if(t*i>DEPTH) mul(b,t),t=1 ; mul(b,t); } /************************************************************************/ /* 组合数 */ /************************************************************************/ void combination(bignum_t a,int m,int n) { int*t=new int[m+1]; memset((void*)t,0,sizeof(int)*(m+1)); comp(t,n+1,m,1); comp(t,2,m-n,-1); convert(t,m,a); delete[]t ; } /************************************************************************/ /* 排列数 */ /************************************************************************/ void permutation(bignum_t a,int m,int n) { int i,t=1 ; memset(a,0,sizeof(bignum_t)); a[0]=a[1]=1 ; for(i=m-n+1;i<=m;t*=i++) if(t*i>DEPTH) mul(a,t),t=1 ; mul(a,t); } #define SGN(x) ((x)>0?1:((x)<0?-1:0)) #define ABS(x) ((x)>0?(x):-(x)) int read(bignum_t a,int&sgn,istream&is=cin) { char str[MAX*DIGIT+2],ch,*buf ; int i,j ; memset((void*)a,0,sizeof(bignum_t)); if(!(is>>str))return 0 ; buf=str,sgn=1 ; if(*buf=='-')sgn=-1,buf++; for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--) ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ; for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0'); for(i=1;i<=a[0];i++) for(a[i]=0,j=0;j<DIGIT;j++) a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ; for(;!a[a[0]]&&a[0]>1;a[0]--); if(a[0]==1&&!a[1])sgn=0 ; return 1 ; } struct bignum { bignum_t num ; int sgn ; public : inline bignum() { memset(num,0,sizeof(bignum_t)); num[0]=1 ; sgn=0 ; } inline int operator!() { return num[0]==1&&!num[1]; } inline bignum&operator=(const bignum&a) { memcpy(num,a.num,sizeof(bignum_t)); sgn=a.sgn ; return*this ; } inline bignum&operator=(const int a) { memset(num,0,sizeof(bignum_t)); num[0]=1 ; sgn=SGN (a); add(num,sgn*a); return*this ; } ; inline bignum&operator+=(const bignum&a) { if(sgn==a.sgn)add(num,a.num); else if (sgn&&a.sgn) { int ret=comp(num,a.num); if(ret>0)sub(num,a.num); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memcpy(num,a.num,sizeof(bignum_t)); sub (num,t); sgn=a.sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if(!sgn) memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ; return*this ; } inline bignum&operator+=(const int a) { if(sgn*a>0)add(num,ABS(a)); else if(sgn&&a) { int ret=comp(num,ABS(a)); if(ret>0)sub(num,ABS(a)); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memset(num,0,sizeof(bignum_t)); num[0]=1 ; add(num,ABS (a)); sgn=-sgn ; sub(num,t); } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if (!sgn)sgn=SGN(a),add(num,ABS(a)); return*this ; } inline bignum operator+(const bignum&a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); ret.sgn=sgn ; ret+=a ; return ret ; } inline bignum operator+(const int a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); ret.sgn=sgn ; ret+=a ; return ret ; } inline bignum&operator-=(const bignum&a) { if(sgn*a.sgn<0)add(num,a.num); else if (sgn&&a.sgn) { int ret=comp(num,a.num); if(ret>0)sub(num,a.num); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memcpy(num,a.num,sizeof(bignum_t)); sub(num,t); sgn=-sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if(!sgn)add (num,a.num),sgn=-a.sgn ; return*this ; } inline bignum&operator-=(const int a) { if(sgn*a<0)add(num,ABS(a)); else if(sgn&&a) { int ret=comp(num,ABS(a)); if(ret>0)sub(num,ABS(a)); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memset(num,0,sizeof(bignum_t)); num[0]=1 ; add(num,ABS(a)); sub(num,t); sgn=-sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if (!sgn)sgn=-SGN(a),add(num,ABS(a)); return*this ; } inline bignum operator-(const bignum&a) { bignum ret ; memcpy(ret.num,num,sizeof(bignum_t)); ret.sgn=sgn ; ret-=a ; return ret ; } inline bignum operator-(const int a) { bignum ret ; memcpy(ret.num,num,sizeof(bignum_t)); ret.sgn=sgn ; ret-=a ; return ret ; } inline bignum&operator*=(const bignum&a) { bignum_t t ; mul(t,num,a.num); memcpy(num,t,sizeof(bignum_t)); sgn*=a.sgn ; return*this ; } inline bignum&operator*=(const int a) { mul(num,ABS(a)); sgn*=SGN(a); return*this ; } inline bignum operator*(const bignum&a) { bignum ret ; mul(ret.num,num,a.num); ret.sgn=sgn*a.sgn ; return ret ; } inline bignum operator*(const int a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); mul(ret.num,ABS(a)); ret.sgn=sgn*SGN(a); return ret ; } inline bignum&operator/=(const bignum&a) { bignum_t t ; div(t,num,a.num); memcpy (num,t,sizeof(bignum_t)); sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ; return*this ; } inline bignum&operator/=(const int a) { int t ; div(num,ABS(a),t); sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a); return*this ; } inline bignum operator/(const bignum&a) { bignum ret ; bignum_t t ; memcpy(t,num,sizeof(bignum_t)); div(ret.num,t,a.num); ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ; return ret ; } inline bignum operator/(const int a) { bignum ret ; int t ; memcpy(ret.num,num,sizeof(bignum_t)); div(ret.num,ABS(a),t); ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a); return ret ; } inline bignum&operator%=(const bignum&a) { bignum_t t ; div(t,num,a.num); if(num[0]==1&&!num[1])sgn=0 ; return*this ; } inline int operator%=(const int a) { int t ; div(num,ABS(a),t); memset(num,0,sizeof (bignum_t)); num[0]=1 ; add(num,t); return t ; } inline bignum operator%(const bignum&a) { bignum ret ; bignum_t t ; memcpy(ret.num,num,sizeof(bignum_t)); div(t,ret.num,a.num); ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ; return ret ; } inline int operator%(const int a) { bignum ret ; int t ; memcpy(ret.num,num,sizeof(bignum_t)); div(ret.num,ABS(a),t); memset(ret.num,0,sizeof(bignum_t)); ret.num[0]=1 ; add(ret.num,t); return t ; } inline bignum&operator++() { *this+=1 ; return*this ; } inline bignum&operator--() { *this-=1 ; return*this ; } ; inline int operator>(const bignum&a) { return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0); } inline int operator>(const int a) { return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0); } inline int operator>=(const bignum&a) { return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0); } inline int operator>=(const int a) { return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0); } inline int operator<(const bignum&a) { return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0); } inline int operator<(const int a) { return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0); } inline int operator<=(const bignum&a) { return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0); } inline int operator<=(const int a) { return sgn<0?(a<0?comp(num,-a)>=0:1): (sgn>0?(a>0?comp(num,a)<=0:0):a>=0); } inline int operator==(const bignum&a) { return(sgn==a.sgn)?!comp(num,a.num):0 ; } inline int operator==(const int a) { return(sgn*a>=0)?!comp(num,ABS(a)):0 ; } inline int operator!=(const bignum&a) { return(sgn==a.sgn)?comp(num,a.num):1 ; } inline int operator!=(const int a) { return(sgn*a>=0)?comp(num,ABS(a)):1 ; } inline int operator[](const int a) { return digit(num,a); } friend inline istream&operator>>(istream&is,bignum&a) { read(a.num,a.sgn,is); return is ; } friend inline ostream&operator<<(ostream&os,const bignum&a) { if(a.sgn<0) os<<'-' ; write(a.num,os); return os ; } friend inline bignum sqrt(const bignum&a) { bignum ret ; bignum_t t ; memcpy(t,a.num,sizeof(bignum_t)); sqrt(ret.num,t); ret.sgn=ret.num[0]!=1||ret.num[1]; return ret ; } friend inline bignum sqrt(const bignum&a,bignum&b) { bignum ret ; memcpy(b.num,a.num,sizeof(bignum_t)); sqrt(ret.num,b.num); ret.sgn=ret.num[0]!=1||ret.num[1]; b.sgn=b.num[0]!=1||ret.num[1]; return ret ; } inline int length() { return :: length(num); } inline int zeronum() { return :: zeronum(num); } inline bignum C(const int m,const int n) { combination(num,m,n); sgn=1 ; return*this ; } inline bignum P(const int m,const int n) { permutation(num,m,n); sgn=1 ; return*this ; } }; bignum a[1005]; void Init() { a[0] = 0; a[1] = 1; for(int i=2; i<1005; i++) a[i] = a[i-1] + a[i-2]; } int main() { Init(); int T, n; scanf("%d",&T); while(T--) { scanf("%d",&n); bignum m; cin>>m; int k; for(int i=0; i<1005; i++) { if(m+2 <= a[i]) { k = i; break; } } bignum tmp; tmp = m - a[k-2]; bignum x; x = 258280327; cout<<tmp%x<<endl; } return 0; }
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