hdu2424 Gary's Calculator 高精度计算

Gary's Calculator

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 857    Accepted Submission(s): 188

Problem Description

Gary has finally decided to find a calculator to avoid making simple calculational mistakes in his math exam. Unable to find a suitable calculator in the market with enough precision, Gary has designed a high-precision calculator himself. Can you help him to write the necessary program that will make his design possible?
For simplicity, you only need to consider two kinds of calculations in your program: addition and multiplication. It is guaranteed that all input numbers to the calculator are non-negative and without leading zeroes.

 

Input

There are multiple test cases in the input file. Each test case starts with one positive integer N (N < 20), followed by a line containing N strings, describing the expression which Gary's calculator should evaluate. Each of the N strings might be a string representing a non-negative integer, a "*", or a "+". No integer in the input will exceed 10⁹.
Input ends with End-of-File.

 

Output

For each test case, please output one single integer (with no leading zeros), the answer to Gary's expression. If Gary's expression is invalid, output "Invalid Expression!" instead. Please use the format indicated in the sample output. 

 

Sample Input

3100 + 600320 * 42+ 500520 + 300 * 20

 

Sample Output

Case 1: 700Case 2: 80Case 3: Invalid Expression!Case 4: 6020

 

Source

2008 Asia Hangzhou Regional Contest Online

 

Recommend

lcy

 这道题用到了高精度计算的加法和乘法，而且这道题根本就不用转化字符串，直接可以拿模板来用，不过拿模板来用
的时候要看清楚模板的精度，这道题自己可能不算难，但对我来说是一次考验，因为这道题有很多条件，我又经常漏条件
所以导致这次设置了好多个变量，导致有时候自己都分不清楚了，做题的的时候先想一些技巧，不要一有想法就做，找一下
自己的思路先大概过一遍，不要让自己以后也这样，代码的风格自己以后都不想看了。
大数乘大数乘法模板

void mult(char a[],char b[],char s[]){    int i,j,k=0,alen,blen,sum=0,res[165][165]={0},flag=0;//通过设置数组可以改变精度这个为165位    char result[165];    alen=strlen(a);blen=strlen(b);     for (i=0;i<alen;i++)for (j=0;j<blen;j++) res[i][j]=(a[i]-'0')*(b[j]-'0');for (i=alen-1;i>=0;i--)        {            for (j=blen-1;j>=0;j--) sum=sum+res[i+blen-j-1][j];            result[k]=sum%10;            k=k+1;            sum=sum/10;        }for (i=blen-2;i>=0;i--)        {            for (j=0;j<=i;j++) sum=sum+res[i-j][j];            result[k]=sum%10;            k=k+1;            sum=sum/10;        }if (sum!=0) {result[k]=sum;k=k+1;}for (i=0;i<k;i++) result[i]+='0';for (i=k-1;i>=0;i--) s[i]=result[k-1-i];s[k]='\0';while(1)        {if (strlen(s)!=strlen(a)&&s[0]=='0') strcpy(s,s+1);elsebreak;        }}

大数加大数模板

void add(char a[],char b[],char back[]){    int i,j,k,up,x,y,z,l;    char *c;    if (strlen(a)>strlen(b)) l=strlen(a)+2; else l=strlen(b)+2;    c=(char *) malloc(l*sizeof(char));    i=strlen(a)-1;    j=strlen(b)-1;    k=0;up=0;    while(i>=0||j>=0){if(i<0) x='0'; else x=a[i];if(j<0) y='0'; else y=b[j];z=x-'0'+y-'0';if(up) z+=1;if(z>9) {up=1;z%=10;} else up=0;c[k++]=z+'0';i--;j--;}    if(up) c[k++]='1';    i=0;    c[k]='\0';    for(k-=1;k>=0;k--)        back[i++]=c[k];    back[i]='\0';} 

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>using namespace std;void mult(char a[],char b[],char s[]){    int i,j,k=0,alen,blen,sum=0,res[165][165]={0},flag=0;    char result[165];    alen=strlen(a);blen=strlen(b);     for (i=0;i<alen;i++)for (j=0;j<blen;j++) res[i][j]=(a[i]-'0')*(b[j]-'0');for (i=alen-1;i>=0;i--)        {            for (j=blen-1;j>=0;j--) sum=sum+res[i+blen-j-1][j];            result[k]=sum%10;            k=k+1;            sum=sum/10;        }for (i=blen-2;i>=0;i--)        {            for (j=0;j<=i;j++) sum=sum+res[i-j][j];            result[k]=sum%10;            k=k+1;            sum=sum/10;        }if (sum!=0) {result[k]=sum;k=k+1;}for (i=0;i<k;i++) result[i]+='0';for (i=k-1;i>=0;i--) s[i]=result[k-1-i];s[k]='\0';while(1)        {if (strlen(s)!=strlen(a)&&s[0]=='0') strcpy(s,s+1);elsebreak;        }}void add(char a[],char b[],char back[]){    int i,j,k,up,x,y,z,l;    char *c;    if (strlen(a)>strlen(b)) l=strlen(a)+2; else l=strlen(b)+2;    c=(char *) malloc(l*sizeof(char));    i=strlen(a)-1;    j=strlen(b)-1;    k=0;up=0;    while(i>=0||j>=0){if(i<0) x='0'; else x=a[i];if(j<0) y='0'; else y=b[j];z=x-'0'+y-'0';if(up) z+=1;if(z>9) {up=1;z%=10;} else up=0;c[k++]=z+'0';i--;j--;}    if(up) c[k++]='1';    i=0;    c[k]='\0';    for(k-=1;k>=0;k--)        back[i++]=c[k];    back[i]='\0';} int flage[100],M,open=0;char num1[100][25],sum[100],temp[100];int main(){int T;M=0;while(scanf("%d",&T)!=EOF){memset(num1,0,sizeof(num1));memset(sum,0,sizeof(sum));memset(flage,0,sizeof(flage));memset(temp,0,sizeof(temp));for(int i=0;i<T;i++){cin>>num1[i];}int dingshitao = 0;open=0;for( i=0;i<T;i++){if(num1[i][0]!='*'&&num1[i][0]!='+'){flage[i]=1;  }if(num1[i][0]=='*'){flage[i]=2;}if(i%2==1&&strlen(num1[i])!=1){dingshitao = 1;}}if(dingshitao == 1){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;continue;}if(T==0){cout<<"Case "<<++M<<": "<<"0"<<endl;continue;}if(T==1){if(flage[0]==1)cout<<"Case "<<++M<<": "<<num1[0]<<endl;else cout<<"Case "<<++M<<": Invalid Expression!"<<endl;continue;}if(T==2){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;continue;}if(T==3&&flage[1]==1){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;continue;}for(i=0;i<T;i++){if(T%2==0){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;open=1;break;}if(i%2==0)//注意技巧的鱼运用{if(flage[i]!=1){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;open=1;break;}}elseif(flage[i]==1){cout<<"Case "<<++M<<": Invalid Expression!"<<endl;open=1;break;}}int f=0;for(i=1;i<T-1;i=i+2){if(flage[i]==2){if(!f){strcpy(temp,num1[i-1]);f=1;}if(flage[i-1]!=3||flage[i+1]!=3){mult(temp,num1[i+1],temp);flage[i-1]=3;flage[i+1]=3;}}else{add(temp,sum,sum);memset(temp,0,sizeof(temp));f=0;}}if(flage[T-2]==2){add(temp,sum,sum);//memset(temp,0,sizeof(temp));}for(i=1;i<T-1;i=i+2){if(flage[i]==0){if((flage[i-1]!=3||flage[i+1]!=3)){if(flage[i-1]==1){add(num1[i-1],sum,sum);flage[i-1]=3;}if(flage[i+1]==1){add(num1[i+1],sum,sum);flage[i+1]=3;}}}}if(!open)//因为open 没有初始化找了好久{if(sum[0]=='0')cout<<"Case "<<++M<<": "<<"0"<<endl;elsecout<<"Case "<<++M<<": "<<sum<<endl;}}return 0;}/*这道题不能来不难的，我没有深入考虑，导致了许多判断，如果一开始有一些设计不会导致这么麻烦，这么多的判断语句，判断语句越多说明错误的的几率越大，所以以后要尽量避免，总有一些情况是你想不到的所以以后要慎用判断，还有就是对一些变量的运用总会忘记初始化，每次用完就要初始化，这不是第一次了这道题最大的错误就是自己没有没有把所有的情况找到就开始用判断。太多的判断自己都不想再看了，发现自己程序不对就设置变量这样越设置越多最后自己都不知道什么意思了*/