PCK本戦：後になって書いてみた編

pck2012_1.c

#include <stdio.h>

enum {A, X, Y, Z, W, B, ERROR};
int map[6][2] = {
	{X, Y},
	{ERROR, Z},
	{X, ERROR},
	{W, B},
	{B, Y},
	{Y, X}
};

int main(void)
{
	int sw;
	int now_pos;
	
	for (now_pos = A; (sw = getchar()) != '#';){
		if (sw == '\n'){
			puts(now_pos == B ? "Yes" : "No");
			now_pos = A;
			continue;
		}
		if (now_pos == ERROR) continue;
		now_pos = map[now_pos][sw == '1'];
	}
	return (0);
}

pck2012_2.c

#include <stdio.h>

#define LIMIT	(10000)

int main(void)
{
	int n;
	int mem[2][10000];
	int ct;
	int under_line;
	int dl;
	int i, j;
	int index;

	while (scanf("%d", &n), n){
		for (i = 0; i < n; i++){
			scanf("%d", &mem[0][i]);
		}

		under_line = n;
		for (ct = 0; ct < LIMIT; ct++){
			for (i = 0; i < under_line && mem[ct & 1][i] == i + 1; i++);
			if (i == under_line) break;

			dl = index = 0;
			for (i = 0; i < under_line; i++){
				dl -= !(--mem[ct & 1][i]);
				if (mem[ct & 1][i]) mem[!(ct & 1)][index++] = mem[ct & 1][i];
			}
			mem[!(ct & 1)][index] = under_line;
			under_line = under_line + 1 + dl;
		}
		printf("%d\n", ct != LIMIT ? ct : -1);
	}

	return (0);
}

pck2012_3.c

#include <stdio.h>

double mpow(double x, int n);
int ch2dec(char hex);
void bit_conv(char *bit, int num);

int main(void)
{
	char bit[64];
	char format[16];
	double frac;
	int dec;
	int n;
	int i;
	int max;
	
	scanf("%d", &n);
	while (n--){
		getchar();
		for (i = 0; i < 8; i++){
			bit_conv(bit + (i * 4), ch2dec(getchar()));
		}
		
		if (*bit) printf("-");
		
		dec = 0;
		for (i = 0; i < 24; i++){
			dec = (dec << 1) + bit[1 + i];
		}
		
		frac = max = 0;
		for (i = 0; i < 7; i++){
			if (bit[25 + i]){
				frac += mpow(0.5, i + 1);
				max = i;
			}
		}
		
		sprintf(format, "%s.%d%s\n", "%", max + 1, "lf");
		
		printf(format, frac + dec);
	}
	
	return (0);
}

void bit_conv(char *bit, int num)
{
	int i;
	int index;
	
	for (index = 0, i = 3; i >= 0; i--, index++){
		bit[index] = (num >> i) & 1;
	}
	return;
}

int ch2dec(char hex){
	if ('0' <= hex && hex <= '9') return (hex - '0');
	return ((hex - 'a') + 10);
}

double mpow(double x, int n)
{
	int i;
	double ret;
	
	ret = 1;
	for (i = 0; i < n; i++){
		ret *= x;
	}
	
	return (ret);
}