pku2295 A DP Problem

题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=2295

题意简述：解方程，并且只有'+'和'-'两种运算符。

解题思路：直接模拟，把未知项的系数和已知项的系数算出来再计算就行了。（这个题目的一个致命trick就是，可能结果是本来是小数，我们要取小于其的最大整数，结果为正时，当然直接取整，如果是负的呢？这里就得注意）

代码（写的非常搓）：

#include<stdio.h>#include<string.h>using namespace std;int main(){ int t; scanf("%d",&t); getchar(); while(t--) { char s[265]; gets(s); int len=strlen(s); int a=-1; for(int i=0;i<len;i++) if(s[i]=='=') { a=i; break; } if(a<=0||a==len-1) { printf("IMPOSSIBLE/n"); continue; } int numx1=0,num1=0; for(int i=0;i<a;i++) { if(s[i]=='x') { if(i==0) { numx1+=1; } else if(s[i-1]-'0'<0||s[i-1]-'0'>9) { if(s[i-1]=='-') numx1-=1; else numx1+=1; } else { int k=i-1; int tt=k; while(k>=0&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(s[k]=='-') sum=-sum; numx1+=sum; } } if(s[i]=='-'||s[i]=='+') { if(i!=0&&s[i-1]!='x') { int k=i-1; int tt=k; while(k>=0&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(k>=0&&s[k]=='-') sum=-sum; num1+=sum; } } } if(s[a-1]!='x') { int k=a-1; int tt=k; while(k>=0&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(k>=0&&s[k]=='-') sum=-sum; num1+=sum; } int numx2=0,num2=0; for(int i=a+1;i<len;i++) { if(s[i]=='x') { if(i==a+1) { numx2+=1; } else if(s[i-1]-'0'<0||s[i-1]-'0'>9) { if(s[i-1]=='-') numx2-=1; else numx2+=1; } else { int k=i-1; int tt=k; while(k>=a+1&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(s[k]=='-') sum=-sum; numx2+=sum; } } if(s[i]=='-'||s[i]=='+') { if(i!=a+1&&s[i-1]!='x') { int k=i-1; int tt=k; while(k>=a+1&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(k>=a+1&&s[k]=='-') sum=-sum; num2+=sum; } } } if(s[len-1]!='x') { int k=len-1; int tt=k; while(k>=a+1&&s[k]-'0'>=0&&s[k--]-'0'<=9); int sum=0; for(int j=k+1;j<=tt;j++) sum=sum*10+s[j]-'0'; if(k>=a+1&&s[k]=='-') sum=-sum; num2+=sum; } if(numx1==numx2&&num1==num2) { printf("IDENTITY/n"); continue; } else if(numx1==numx2&&num1!=num2) { printf("IMPOSSIBLE/n"); continue; } else { int aa; double ans=double(num2-num1)/(numx1-numx2); if(ans>=0.0) aa=(int)ans; else aa=(int)ans-1; printf("%d/n",aa); } } return 0;}