"Algorithmic Learning Notes" 35. High-precision vertical multiplication SJTU OJ 1274

Description

Enter A, b

The vertical multiplication of the output a*b is shown in the sample format.

Sample INPUT1

119

Sample OUTPUT1

11 9--99

Sample Input2

1010

Sample Output2

    10    10   ---   100

Sample INPUT3

101101

Sample OUTPUT3

        101        101      -----        101      101      -----      10201

Sample Input4

100862

Sample Output4

        2    10086    -----       12      16    2    -----    20172

Sample INPUT5

1234001238412039847102934

Sample OUTPUT5

              12340012384           12039847102934 ------------------------              49360049536             37020037152           111060111456           24680024768         12340012384        86380086688       49360049536      98720099072    111060111456   37020037152  24680024768 12340012384 ------------------------ 148571862351672082734656


Although the problem is not algorithmic difficulty, but one of the inspiration is to make a thorough study of the sample, such as these cases have hidden several conditions.
1. If a number is 1 digits, and the number of digits of the other is >=3, the short one is placed on it, and if 2 or 1 does not need to be processed
2. If only one row in the result is output directly
3. If a row in the result is all 0, do not output

The code is as follows

#include <iostream>#include<cstring>using namespacestd;Chara_input[ -]={0};Charb_input[ -]={0};Charres[ -][ -]={0};//store calculation results for each multiplicationintfinal_res[ -]={0};inlinevoidPrintblank (intNCharFlag =' '){     for(inti =0; I < n; ++i) {cout<<Flag; }    return;}voidCalCharA[],CharB[],intA_len,intB_len) {    intFinal_res_len =0;  for(inti = b_len-1; I >=0; I.)//B[i] is one of the bottom numbers    {        intremain =0;//stores the number of digits to be rounded        intindex = b_len-1I//The results of this multiplication are stored in Res[index]//Reverse Storage        intCur =0;  for(intj = A_len-1; J >=0; --j) {intTmp_res = (a[j]-'0') * (b[i]-'0') +remain; Res[index][cur]= (tmp_res)%Ten+'0'; Remain= Tmp_res/Ten; Cur++; }        if(remain>0)//in the end, one more.Res[index][cur] = remain +'0'; }    //calculate the final result     for(inti =0; i < B_len; ++i) {intremain =0;//the number to be rounded         for(intj =0; J < strlen (Res[i]); ++J)//the lowest I bit remains the same        {            //Final_res J-bit and j-i-1-bit alignment            intTemp_res = Final_res[j+i] + (res[i][j]-'0') +remain; Final_res[j+i] = temp_res%Ten; Remain= Temp_res/Ten; }        if(remain>0)//and the final rounding .{Final_res[strlen (res[i])+i] =remain; if(i==b_len-1) Final_res_len= strlen (Res[i]) +i+1; }Else{            if(i==b_len-1) Final_res_len= strlen (Res[i]) +i; }} printblank (Final_res_len-A_len); cout<<a<<Endl; Printblank (Final_res_len-B_len); cout<<b<<Endl; Printblank (Final_res_len,'-'); cout<<Endl; BOOLTooutput =false;intCot =0;  for(inti =0; i < B_len; ++i)if(b[b_len-1-i]!='0') Cot++; Tooutput= cot>1; if(tooutput) { for(inti =0; i < B_len; ++i)if(b[b_len-1-i]!='0') {Printblank (Final_res_len-Strlen (Res[i])-i);  for(intj = strlen (Res[i])-1; j>=0; --j) cout<<Res[i][j]; cout<<Endl; //Cout<<strlen (res[i]) << ":" <<res[i]<<endl;} printblank (Final_res_len,'-'); cout<<Endl; }     for(inti = final_res_len-1; I >=0; --i) {cout<<Final_res[i]; }cout<<Endl;}intMainintargcChar Const*argv[]) {    Chara[ -]={0}; Charb[ -]={0}; CIN>>A_input; CIN>>B_input; intA_len =0; intB_len =0; //Remove Front 0    BOOLbegan =false;  for(inti =0; I < strlen (a_input); ++i) {if(!began and a_input[i]!='0') began=true; if(began) A[a_len++] =A_input[i]; } began=false;  for(inti =0; I < strlen (b_input); ++i) {if(!began and b_input[i]!='0') began=true; if(began) B[b_len++] =B_input[i]; }        //the above is the initialization of    if(a_len>=3and B_len = =1) Cal (B,a,b_len,a_len); Elsecal (A,b,a_len,b_len); return 0;}/*12340012384 12039847102934------------------------49360049536 370 20037152 111060111456 24680024768 12340012384 86380086688 49360049536 987200 99072 111060111456 37020037152 24680024768 12340012384------------------------148571862351672082734656*/

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"Algorithmic Learning Notes" 35. High-precision vertical multiplication SJTU OJ 1274