joj 1920解題報告

 

 1920: Jojer

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	733	231	Standard

Mr Jojer is a teacher, and he is crazy in programming, he wants the computer to do every thing for him, one day, one of his colleague Mr NEtiger asks him to write a procedure for him. But Mr Jojer has another difficult problem, so he asks you to do it for him. Do you want a try? Ok, let's begin!

There are two numbers for the pupils to operate, Mr NEtiger wants to know the answers of plus, minus, multiply, divide,so he can check the answers which the pupils hand in.

The input contain two numbers, you should output the answers in the order of plus, minus, multiply, divide. You may assume that they are dividable and no remainder.

Is it easy? But you should know the numbers may as long as 50 digits.

Sample Input

10 2
15 3

Sample Output

12
8
20
5
18
12
45
5

水題一道，就是高精度計算，就是考察高精度四則運算。

代碼：

語言：c++

#include<iostream> #include<cstring> #define N 1001 using namespace std; int main() { char m1[100],m2[100],c1[100],c2[100]; char *addition(char *m1,char *m2); char *subtract(char *m1,char *m2); char *multiply(char *m1,char *m2); void high_precise_division(char *c1,char *c2); while(cin>>m1>>m2) { strcpy(c1,m1); strcpy(c2,m2); high_precise_division(c1,c2); cout<<addition(m1,m2)<<endl<<subtract(m1,m2)<<endl<<multiply(m1,m2)<<endl<<c1<<endl; } return 0; } char *addition(char *m1,char *m2) { int *str2int(char *str); char *int2str(int *a,int n); int check(int *a,int n); int len1=strlen(m1),len2=strlen(m2),len,i,*t1,*t2; len=(len1>=len2)?len1:len2; t1=str2int(m1); t2=str2int(m2); for(i=len1;i<=len;++i) t1[i]=0; for(i=len2;i<=len;++i) t2[i]=0; for(i=0;i<len;++i) t1[i]+=t2[i]; len=check(t1,len); return int2str(t1,len); } int *str2int(char *str) { int len=strlen(str); int *a=new int[(len+1)*sizeof(int)]; for(int i=0;i<len;++i) a[i]=(int)str[len-1-i]-'0'; return a; } char *int2str(int *a,int n) { int len=n; char *str=new char[(len+1)*sizeof(char)]; for(int i=0;i<len;++i) str[i]=(char)a[len-1-i]+'0'; str[len]='/0'; return str; } int check(int *a,int n) { int len=n,i; while(len>1&&a[len-1]==0) --len; for(i=0;i<len;++i) if(a[i]>=10) { a[i+1]+=a[i]/10; a[i]%=10; } if(a[i]!=0) len=i+1; return len; } char *subtract(char *m1,char *m2) { int *str2int(char *str); int len1=strlen(m1),len2=strlen(m2),len,i,c,cf=0,begin,*t1,*t2,n; char *temp,*csub; len=(len1>=len2)?len1:len2; temp=new char[(len+2)*sizeof(char)]; csub=new char[(len+3)*sizeof(char)]; if(len>len1||(len1==len2&&strcmp(m1,m2)<0)) { cf=1; strcpy(temp,m1); strcpy(m1,m2); strcpy(m2,temp); len1=strlen(m1); len2=strlen(m2); } t1=str2int(m1); t2=str2int(m2); for(i=len1;i<len;++i) t1[i]=0; for(i=len2;i<len;++i) t2[i]=0; for(i=0;i<len;++i) { if(t1[i]>=t2[i]) t1[i]-=t2[i]; else { t1[i]-=t2[i]; c=(t1[i]*(-1)/10+1); t1[i]+=c*10; t1[i+1]-=c; } } n=i; begin=0; while(n>1&&t1[n-1]==0) --n; if(cf==1) { ++n; begin=1; csub[0]='-'; } for(i=begin;i<n;++i) csub[i]=(char)t1[n-1-i]+'0'; csub[n]='/0'; return csub; } char *multiply(char *m1,char *m2) { int *str2int(char *str); char *int2str(int *a,int n); int check(int *a,int n); int len1=strlen(m1),len2=strlen(m2),len,*t1,*t2,*prod,i,j; len=len1+len2; t1=str2int(m1); t2=str2int(m2); prod=new int[(len+1)*sizeof(int)]; for(i=0;i<=len;++i) prod[i]=0; for(i=0;i<len1;++i) for(j=0;j<len2;++j) prod[i+j]+=t1[i]*t2[j]; len=check(prod,len); return int2str(prod,len); } int compare(char * c1,int s1,int e1,char *c2,int s2,int e2)/*大整數比較大小*/ { while(c1[s1]=='0'&&s1<e1)s1++; while(c2[s2]=='0'&&s2<e2)s2++; if(e1-s1>e2-s2)return 1; if(e1-s1<e2-s2)return -1; int i1;int f11=0; for(i1=0;i1<=e1-s1;i1++) { if(c1[s1+i1]>c2[s2+i1]){f11=1;break;} if(c2[s2+i1]>c1[s1+i1])break; } if(i1>e1-s1)return 0; if(f11)return 1; return -1; } void sub(char *c1,int l1,char *c2,int n,int l)/*兩個大整數相減,結果放在c1中*/ { int fig=0; if(l1>l) { fig=1; int i5; for(i5=l;i5>0;i5--) c2[i5]=c2[i5-1]; c2[0]='0';l++; } int jw=0; int i4; char tc[N]; for(i4=l-1;i4>=0;i4--) { tc[i4]=((c2[i4]-48)*n+jw)%10+48; jw=((c2[i4]-48)*n+jw)/10; } jw=0; for(i4=l-1;i4>=0;i4--) { int ttt=jw; if(c1[i4]-tc[i4]-jw>=0) { tc[i4]=c1[i4]-tc[i4]-jw+48; jw=0; } else { jw=(tc[i4]-c1[i4]+jw); if(jw%10==0)jw/=10; else jw=jw/10+1; tc[i4]=jw*10+c1[i4]-tc[i4]-ttt+48; } } if(fig) { for(i4=0;i4<l-1;i4++)c2[i4]=c2[i4+1]; l--; c2[l]='/0'; } tc[l1]='/0'; for(i4=0;i4<=l1;i4++) c1[i4]=tc[i4]; } void high_precise_division(char *c1,char *c2) { int len1,len2; len1=strlen(c1); len2=strlen(c2); int i,j,k,ip; i=0; while(c1[i]=='0'&&i<len1-1)i++; for(j=0;i<len1;j++,i++)c1[j]=c1[i]; len1=j;c1[j]='/0'; i=0; while(c2[i]=='0'&&i<len2-1)i++; for(j=0;i<len2;j++,i++)c2[j]=c2[i]; len2=j;c2[j]='/0'; /*while(c1[0]=='0'&&i<len1-1) { for(k=0;k<len1-1;k++) c1[k]=c1[k+1]; len1--; }去掉前面的0*/ /*while(c2[0]=='0'&&j<len2-1) { for(k=0;k<len2-1;k++) c2[k]=c2[k+1]; len2--; }去掉前面的0*/ c1[len1]='/0'; c2[len2]='/0'; if(strcmp(c2,"0")==0)return ;/*當除數爲0時*/ if(strcmp(c1,"0")==0){strcpy(c2,"0");return ;}/*當被除數爲0時*/ if(len1<len2||len1==len2&&compare(c1,0,len1-1,c2,0,len2-1)<0) { strcpy(c2,c1); c1[0]='0';c1[1]='/0'; return ; } else { ip=0; char product[N],*pr;/*部分積*/ pr=product; for(ip=0;ip<len2-1;ip++) pr[ip]=c1[ip]; for(i=0;i<=len1-len2;i++) { pr[ip++]=c1[len2-1+i]; if(ip>=len2 && compare(pr,0,ip-1,c2,0,len2-1)>=0) { char tc[N]; for(j=1;j<=9;j++) { for(k=0;k<ip;k++)tc[k]=pr[k]; sub(tc,ip,c2,j,len2);/*pr-c2*j結果放在pr中*/ for(k=0;tc[k]!='/0';k++); if(compare(tc,0,k-1,c2,0,len2-1)<0) break; } strcpy(pr,tc); ip=strlen(pr); c1[i]=j+48; while(pr[0]=='0'&&ip>1) { for(j=0;j<ip-1;j++) pr[j]=pr[j+1]; ip--; } if(ip==1&&pr[0]=='0')ip--; } else c1[i]='0'; } while(c1[0]=='0') { for(j=0;j<i-1;j++) c1[j]=c1[j+1]; i--; } c1[i]='/0'; if(ip==0){pr[0]='0';pr[1]='/0';ip=1;} else { while(pr[0]=='0'&&ip>1) { for(j=0;j<ip-1;j++)pr[j]=pr[j+1]; ip--; } } for(j=0;j<ip;j++) c2[j]=pr[j]; c2[ip]='/0'; return ; } }