Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In this problem, you are to solve a very easy linear equation with only one variable x with no parentheses! An example of such equations is like the following:

An expression in its general form, will contain a `=' character with two expressions on its sides. Each expression is made up of one or more terms combined by `+' or `-' operators. No unary plus or minus operators are allowed in the expressions. Each term is either a single integer, or an integer followed by the lower-case character x or the single character x which is equivalent to 1x.

You are to write a program to find the value of x that satisfies the equation. Note that it is possible for the equation to have no solution or have infinitely many. Your program must detect these cases too.

Input 

The first number in the input line, t ( 1t10) is the number of test cases, followed by t lines of length at most 255 each containing an equation. There is no blank character in the equations and the variable is always represented by the lower-case character `x'. The coefficients are integers in the range (0..1000) inclusive.

Output 

The output contains one line per test case containing the solution of the equation. If s is the solution to the equation, the output line should contain s (the ``floor" of s, i.e., the largest integer number less than or equal to s). The output should be `IMPOSSIBLE' or `IDENTITY' if the equation has no solution or has infinite solutions, respectively. Note that the output is case-sensitive.

Sample Input 

2 
2x-4+5x+300=98x
x+2=2+x

Sample Output 

3
IDENTITY

---

#include <stdio.h>
#include <math.h>
void parse(char *s, int& A, int& B) {
    A = 0, B = 0;
    int i, g = 0, f = 0, neg = 1;
    for(i = 0; s[i]; i++) {
        if(s[i] == 'x') {
            if(g)
                A += neg*f;
            else
                A += neg;
            g = 0, f = 0, neg = 1;
        } else {
            if(s[i] == '-') {
                if(g)
                    B += neg*f;
                g = 0, f = 0;
                neg = -1;
            } else if(s[i] == '+') {
                if(g)
                    B += neg*f;
                g = 0, f = 0;
                neg = 1;
            } else
                f = f*10 + s[i]-'0', g = 1;
        }
    }
    if(g)
        B += neg*f;
}
int main() {
    int t, i;
    scanf("%d", &t);
    char s1[502], *s2;
    while(t--) {
        scanf("%s", s1);
        for(i = 0; s1[i]; i++) {
            if(s1[i] == '=') {
                s2 = s1+i+1;
                s1[i] = '\0';
                break;
            }
        }
        int la, lb, ra, rb;
        parse(s1, la, lb);
        parse(s2, ra, rb);
        if(la == ra && lb == rb)
            puts("IDENTITY");
        else if(la == ra && lb != rb)
            puts("IMPOSSIBLE");
        else
            printf("%d\n", (int)floor((double)(rb-lb)/(la-ra)));
    }
    return 0;
}