Equation Solver Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 427   Accepted: 266 Description Write a program that can solve linear equations with one variable. Input The input will contain a number of equations, each one on a separate line. All equations are strings of less than 100 characters which strictly adhere to the following grammar (given in EBNF):
Equation := Expression '=' Expression
Expression := Term { ('+' | '-') Term }
Term := Factor { '*' Factor }
Factor := Number | 'x' | '(' Expression ')'
Number := Digit | Digit Number
Digit := '0' | '1' | ... | '9'
Although the grammar would allow to construct non-linear equations like "x*x=25", we guarantee that all equations occuring in the input file will be linear in x. We further guarantee that all sub-expressions of an equation will be linear in x too. That means, there won't be test cases like x*x-x*x+x=0 which is a linear equation but contains non-linear sub-expressions (x*x).
Note that all numbers occuring in the input are non-negative integers, while the solution for x is a real number. Output For each test case, print a line saying "Equation #i (where i is the number of the test case) and a line with one of the following answers:

• If the equation has no solution, print "No solution.".
• If the equation has infinitely many solutions, print "Infinitely many solutions.".
• If the equation has exactly one solution, print "x = solution" where solution is replaced by the appropriate real number (printed to six decimals).

Print a blank line after each test case.
Sample Input x+x+x=10 4*x+2=19 3*x=3*x+1+2+3 (42-6*7)*x=2*5-10 Sample Output Equation #1 x = 3.333333 Equation #2 x = 4.250000 Equation #3 No solution. Equation #4 Infinitely many solutions. 题意:解一元一次方程 思路:设置两个整型,一个保存当前得到的常数,另一个保存X的系数,然后递归求解..... #include<iostream> using namespace std; char in[200],a[200]; int pos; void cal(int &NUM,int &XISHU) { int nowNUM=0,nowXISHU=0,sign=1; for(;pos<strlen(in);pos++) if(in[pos]>='0'&&in[pos]<='9') nowNUM=nowNUM*10+in[pos]-48; else if(in[pos]=='x') { if(nowNUM) { nowXISHU=nowNUM; nowNUM=0; } else nowXISHU=1; } else if (in[pos]=='(') { pos++; cal(nowNUM,nowXISHU); } else if(in[pos]==')') { nowNUM*=sign; NUM+=nowNUM; XISHU+=nowXISHU; return ; } else if(in[pos]=='+') { NUM+=nowNUM*sign; XISHU+=nowXISHU*sign; nowXISHU=0; nowNUM=0; sign=1; } else if(in[pos]=='-') { NUM+=nowNUM*sign; XISHU+=nowXISHU*sign; nowXISHU=0; nowNUM=0; sign=-1; } else if(in[pos]=='*') { nowNUM*=sign; nowXISHU*=sign; sign=1; pos++; for(int nnum=0,nxishu=0;pos<strlen(in)+1;pos++) if(in[pos]=='(') { pos++; cal(nnum,nxishu); } else if(in[pos]>='0'&&in[pos]<='9') nnum=nnum*10+in[pos]-48; else if(in[pos]=='x') { nxishu=1; } else { if(nxishu) nowXISHU=nowNUM*nxishu; else nowXISHU=nowXISHU*nnum; nowNUM*=nnum; pos--; break; } } NUM+=nowNUM*sign; XISHU+=nowXISHU*sign; return ; } int main() { int t=0; while(cin>>a&&++t) { int i; for(i=0;a[i]!='=';i++) in[i]=a[i]; in[i]='/0'; int NUM=0,XISHU=0,NUM1=0,XISHU1=0; pos=0; cal(NUM,XISHU); strcpy(in,&a[i+1]); pos=0; cal(NUM1,XISHU1); XISHU-=XISHU1; NUM1-=NUM; if(XISHU==0&&NUM1!=0) cout<<"Equation #"<<t<<endl<<"No solution."<<endl<<endl; else if(XISHU==0&&NUM1==0) cout<<"Equation #"<<t<<endl<<"Infinitely many solutions."<<endl<<endl; else { printf("Equation #%d/nx = %.6lf/n/n",t,double(NUM1)/double(XISHU)); } } }