JZOJ · expression evaluates expr [simulation]

Description–

Given a containing only addition and multiplication arithmetic expression, the value of your program to evaluate the expression.

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Input–

Input file expr.in .
Input only one row, your expression is required for the calculation, the expression contains only numbers, the addition operator "+" and multiplication
method the operator "*", and without brackets, all numbers involved in computing are 0 to 231 - an integer between 1.
Input data to ensure that this line only 0 to 9, +, * These 12 characters.

Output–

Output file name expr.out.
Output only one line containing an integer that represents the value of the expression.
Note: When the answer to the length of more than 4, only the last four output request, leading 0 is not output.

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Sample Input–

expr.in1
1+1*3+4

expr.in2
1 + 1234567890 * 1

expr.in3
1+1000000003*1

Sample Output–

expr.out1
8

expr.out2
7891

expr.out3
4

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Notes -

【data range】

For 30% of the data, 0≤ expression addition operator and the multiplication operator the total number of ≦ 100;

For 80% of data, 0≤ expression addition operator and the multiplication operator Total ≤1000;

For 100% of the data, 0≤ expression Total ≤100000 addition operator and the multiplication operator.

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Problem-solving ideas -

For each number appears taken only after its 4

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Code -

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int t,fy,a[100005];
char b[100005];
string s;
int main()
{
	//freopen("expr.in","r",stdin);
	//freopen("expr.out","w",stdout);
	
	cin>>s;
	for (int i=0;i<s.size();++i)
	  if (s[i]<='9' && s[i]>='0')
	    fy=fy*10+int(s[i]-48);
	  else
	    a[++t]=fy%10000,fy=0,b[t]=s[i];
	a[++t]=fy%10000;
	for (int i=1;i<=t;++i)
	  if (b[i]=='*')
		a[i+1]=(a[i+1]*a[i])%10000,a[i]=0;//乘法
    for (int i=2;i<=t;++i)
      a[1]=(a[1]+a[i])%10000;//加法
    printf("%d",a[1]);
	
	return 0;
}