hdu 1002 A + B Problem II 万能大数模板再次成功应用
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A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 122283 Accepted Submission(s): 23433
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
万能模板 很长很长 但是能实现很多很多功能 有些功能我们用不到 可以不进行调用 那么就不会影响我们写的程序的速度
所以是值得推广的
#include <iostream>#include <cstring>#include<math.h>using namespace std;#define DIGIT4 //四位隔开,即万进制#define DEPTH10000 //万进制#define MAX 2000typedef 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 ; }};/*int main(){ bignum a,b,c;cin>>a>>b;cout<<"加法:"<<a+b<<endl;cout<<"减法:"<<a-b<<endl;cout<<"乘法:"<<a*b<<endl;cout<<"除法:"<<a/b<<endl;c=sqrt(a);cout<<"平方根:"<<c<<endl;cout<<"a的长度:"<<a.length()<<endl;cout<<"a的末尾0个数:"<<a.zeronum()<<endl<<endl;cout<<"组合: 从10个不同元素取3个元素组合的所有可能性为"<<c.C(10,3)<<endl;cout<<"排列: 从10个不同元素取3个元素排列的所有可能性为"<<c.P(10,3)<<endl;return 0 ;}*/ //////////////////////////////////////////////////////////////*上面是一个完整的大数模板 已经功能的演示 我只是在下面修改了主函数和加入了 get_prime */int main(){ bignum a,b,c;int cas,t;//while(!cin.eof()) 这句话 不用加 直接while(cin>>...)根本不用加eof //{t=0;cin>>cas; while(cas--) {cin>>a>>b; cout<<"Case "<<++t<<":"<<endl; cout<<a<<" + "<<b <<" = "<<a+b<<endl; if(cas!=0) cout<<""<<endl; //<<"" 这个就代表一个换行符了 以前没有用过cout 长见识了}//} return 0 ;}
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