请用C语言编程判断是谁做的好事，A说：不是我。B说：是C。C说:是D。D说：C胡说，编程判断谁做了好事

在C语言中，这个问题可以通过创建一个简单的结构体或者枚举来表示人物，并使用循环和条件语句来模拟他们的对话。由于C语言本身并没有直接支持类似“聊天”的功能，我们可以编写一个简单的程序来代表这个逻辑。这里假设我们有四个字符`A`, `B`, `C`, 和 `D`，并创建一个变量`culprit` 来存储做好事的人。

#include <stdio.h>

typedef enum {NO_CULPRIT, A, B, C, D} Person;

void hearStatement(Person speaker, Person culprit) {
    switch (speaker) {
        case A:
            if (culprit == C || culprit == D) {
                printf("A says: It's either C or D.\n");
            } else {
                printf("A says: Not me.\n");
            }
            break;
        case B:
            if (culprit == C) {
                printf("B says: It is C.\n");
            } else {
                printf("B's statement doesn't add to the information.\n");
            }
            break;
        case C:
            printf("C says: It's D.\n");
            break;
        case D:
            printf("D says: C is lying!\n");
            // Since we know D is not saying it was themselves, it must be C.
            culprit = C;
            break;
        default:
            printf("Invalid input.\n");
    }
}

int main() {
    Person culprit = NO_CULPRIT; // Initially no one is known to have done the good deed

    // Let's start with the statements and update the culprit accordingly
    hearStatement(A, culprit);
    hearStatement(B, culprit);
    hearStatement(C, culprit);
    hearStatement(D, culprit);

    // Now, output the final result
    if (culprit == A) {
        printf("It turns out that A did the good deed.\n");
    } else if (culprit == B) {
        printf("B did the good deed.\n");
    } else if (culprit == C) {
        printf("C did the good deed.\n");
    } else if (culprit == D) {
        printf("D did the good deed.\n");
    } else {
        printf("No one did the good deed according to this sequence of events.\n");
    }

    return 0;
}