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

Background

Although testing is a technique which does not allow us to prove the correctness of a program, it is used to assess its reliability, thus increasing our confidence in its correctness. In the literature, there are many testing methods to select test cases which are applied as an input to the program.

The Problem

A programmer has the intention of testing a program. With this aim in mind, he needs to know how many test cases he should obtain such that each distinct path of execution is covered with a test case. Your task is to write a program which computes the number of different paths of execution from the initial sentence to the final sentence, in a given program P.

The Input

The first line of the input contains the number of programs. Subsequently, each program is listed. Each program consists of a set of sentences with a keyword per line. The program finishes with the keyword ENDPROGRAM

There are only two types of sentences: conditional and non-conditional. A non-conditional sentence is represented by means of the keyword S. A conditional sentence is represented by the keywords IF, ELSE and END_IF. The condition of the IF sentence is omitted (i.e., only the IF keyword appears) and it is supposed that all IF sentences have a corresponding ELSE. So, for example, the shortest program with a conditional sentence would be:

IF
ELSE
END_IF
ENDPROGRAM

which has 2 paths of execution.

The Output

For every program you are to print a line containing the number N of different paths of execution. It is supposed that N is never greater than 2^32.

Sample Input

4
IF
ELSE
END_IF
ENDPROGRAM
IF
S
ELSE
END_IF
IF
ELSE
S
END_IF
ENDPROGRAM
S
S
S
ENDPROGRAM
S
IF
S
S
ELSE
IF
IF
S
ELSE
S
END_IF
S
ELSE
S
END_IF
END_IF
S
ENDPROGRAM

Sample Output

2
4
1
4

---

題目描述：
給定 IF, ELSE, END_IF 指令，問程式流程的路徑有多少種？

題目解法：
獨立的巢狀迴圈在同一 BLOCK 中，方法數是相乘起來的。
然而 IF 對應的 ELSE 區塊的方法數是相加起來的。
每個 IF 一定存在對應的 ELSE，其實 S 指令是可以被忽略的，並沒有太多功用。

抓 BUG 抓到死，一直 WA。

#include <stdio.h>
#include <string.h>
char stmt[10005][100];
int n, i;
long long calc() {
    long long ret = 0, tmp = 0;
    if(!strcmp(stmt[i], "IF")) {
        i++;
        tmp = 1;
        while(true) {
            if(!strcmp(stmt[i], "IF"))
                tmp *= calc();
            else
                break;
        }
        ret += tmp;
    }
    if(!strcmp(stmt[i], "ELSE")) {
        i++;
        tmp = 1;
        while(true) {
            if(!strcmp(stmt[i], "IF"))
                tmp *= calc();
            else
                break;
        }
        ret += tmp;
    }
    i++;
    return ret;
}
int main() {
    int testcase;
    scanf("%d", &testcase);
    while(testcase--) {
        n = 0;
        while(scanf("%s", &stmt[n]) == 1) {
            if(!strcmp(stmt[n], "ENDPROGRAM"))
                break;
            if(!strcmp(stmt[n], "IF"))
                n++;
            else if(!strcmp(stmt[n], "ELSE"))
                n++;
            else if(!strcmp(stmt[n], "END_IF"))
                n++;
        }
        n++;
        i = 0;
        long long ret = 1;
        while(i < n-1)
            ret *= calc();
        printf("%lld\n", ret);
    }
    return 0;
}