注意幾個問題：
1，分呣是1的時猴不输出
2，分呣小于洞的時猴分呣分子取反
3，注意区分数中的负號和式子的减號
4，好像没有超int范围的test case
input:
231/2+235/2
output:
233

input:
1024/2-512/1
output:
0

input:
-233/2333-233/-2333
output:
0

input:
1/-5--1/7
output:
-2/35

附代妈：
#include <stdio.h>
#include <string.h>
#include <cmath>
using namespace std;
int gcd(int a,int b){
	a=abs(a+0.0),b=abs(b+0.0);
	if(!a)return b;
	if(!b)return a;
	if(a<b)return gcd(a,b%a);
	return gcd(b,a%b);
}
int main() {
	char exp[233];
	while(scanf("%s",exp)>0){
		int l = strlen(exp);
		char op;
		int st = 0;
		int s1;
		int a=0,b=0,c=0,d=0;
		bool fa=0,fb=0,fc=0,fd=0;
		for(int i = 0; i < l; i++){
			switch(st){
				case 0:
					if(exp[i]=='/') {
						st = 1;
						s1=i;
					}
					else if(exp[i]=='-') fa=1;
					else{
						a*=10;
						a+=exp[i]-'0';
					}
					break;
				case 1:
					if(exp[i]=='+'||(i>s1+1 && exp[i]=='-')){
						op = exp[i];
						st = 2;
					}
					else if(exp[i]=='-') fb=1;
					else{
						b*=10;
						b+=exp[i]-'0';
					}
					break;
				case 2:
					if(exp[i]=='/') st = 3;
					else if(exp[i]=='-') fc=1;
					else{
						c*=10;
						c+=exp[i]-'0';
					}
					break;
				case 3:
					if(exp[i]=='-') fd=1;
					else{
						d*=10;
						d+=exp[i]-'0';
					}
					break;
				default:
					break;
			}
		}
		if(fa)a=-a;
		if(fb)b=-b;
		if(fc)c=-c;
		if(fd)d=-d;
		int fm=b*d,fz=a*d+b*c*((op=='+')?1:-1),g=gcd(fm,fz);
		fm/=g,fz/=g;
		if(fm>>31){
			fm=-fm;
			fz=-fz;
		}
		if(!fz)printf("0\n");
		else if(!(fm-1))printf("%d\n",fz);
		else printf("%d/%d\n",fz,fm);
	}
	return 0;
}

• Re:很坑，給幾組樣例吧，這幾組過了應該就木問題了 (27) xgcxgc 2018-07-31 15:08:54