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.
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.
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
3 100 + 600 3 20 * 4 2 + 500 5 20 + 300 * 20
Sample Output
Case 1: 700 Case 2: 80 Case 3: Invalid Expression! Case 4: 6020
Source
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lcy
Statistic | Submit | Discuss | Note
这道题是我第一次套用模板。
用了两个模板。
高精度乘法，和高精度加法。。
本来这道题挺简单的，但是让我漏了几个条件，wa了十几次。。
还有看了一些前辈的blog，感到用java就能十分钟做出来。因为java已经有高精度了。。
#include<stdio.h>
#include<string.h>
#include<malloc.h>
int n;
char ch[15];
void mult(char a[],char b[],char s[])
{
int i,j,k=0,alen,blen,sum=0,res[150][150]={0},flag=0;
char result[150];
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);
else
break;
}
}
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 main()
{
int flag,flag2,ca = 1;
char pre1[102],pre2[102],b[102];
while(scanf("%d",&n)!=EOF)
{
flag = 1,flag2 = 1;
memset(pre1,'\0',sizeof(pre1));
memset(pre2,'\0',sizeof(pre2));
for(int i=1;i<=n;i++)
{
scanf("%s",ch);
if(n%2==0) //一直漏了这个条件，wa呀wa
flag=0;
if(flag)
{
if(i%2==1 )
{
if(ch[0]=='+'||ch[0]=='*')//还有这个
{
flag = 0;
continue;
}
if(flag2==1)
{
memset(b,'\0',sizeof(b));
add(pre1,pre2,b);
strcpy(pre1,b);
strcpy(pre2,ch);
}
else
{
memset(b,'\0',sizeof(b));
mult(pre2,ch,b);
strcpy(pre2,b);
}
}
else
{
if((ch[0]!='+' && ch[0]!='*') || strlen(ch)!=1)
{
flag = 0;
continue;
/*printf("Case %d: Invalid Expression!\n",ca++);
break;*/
}
if(ch[0]=='+')
flag2 = 1; //1代表加法
else
flag2 = 2; //2代表乘法
}
}
}
add(pre1,pre2,b);
if(!flag )
printf("Case %d: Invalid Expression!\n",ca++);
else
{
printf("Case %d: ",ca++);
int i=0;
while(b[i]=='0')
i++;
if(i==strlen(b))
printf("0");
for(;i<strlen(b);i++)
printf("%c",b[i]);
printf("\n");
}
}
return 0;
}