P20@C

<pre name="code" class="cpp">/*
* P20.c
*
* Created on: 2011-10-29
* Author: ZHAO Jing P20 in C
* 程序設計基礎附錄20題參考解答,僅供參考，歡迎討論和PK
*/
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<time.h>
#include<ctype.h>
#include<string.h>
#define EPS 1E-8
#define TRUE 1
#define FALSE 0
#define NEWLINE putchar('\n')
#define READ(x) scanf("%d",&x)
#define WRITELN(x) printf("%d\n",x)
#define ABS(x) (((x)>0)?(x):(-(x))) //絕對值
#define MAX(x,y) ((x)>(y)?(x):(y)) //最大值
#define MIN(x,y) ((x)<(y)?(x):(y)) //最小值
#define SQUARE(x) ((x)*(x)) //平方
#define IMPLY(x,y) (!(x)||(y)) /*邏輯蘊含x->y的定義*/
#define FEQUAL(x,y) ABS((x)-(y))<=EPS //浮點相等
#define SIGN(x) (((x)>0)?1:((x)<0)?-1:0) //返回一個表達式的符號
typedef long long LL;
typedef long double LD;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef char BYTE;
typedef char BOOL;
/**
* 用於qsort的洗牌算子
*/
int compRand(const void* a, const void* b) {
return rand() % 3 - 1;
}
/**
* 作用是將數組前n個元素洗亂產生隨機順序
*/
void shuffle(int *a, int n) {
qsort(a, n, sizeof(int), compRand);
}
/**
* 數組打印程序,打印輸出數組的前n個元素
*/
void printArr(int *a, int n) {
int i;
for (i = 0; i < n; i++)
printf("%d ", a[i]);
NEWLINE;
}
/**
* 變量交換，C中必須使用指針才能達到間接修改變量的效果
*/
void swap(int* a, int *b) {
int t = *a;
*a = *b;
*b = t;
}
/**
* 利用rand()函數產生[start,end]之間的隨機數
*/
int doss(int start, int end) {
if (start > end)
swap(&start, &end);
return rand() % (end - start + 1) + start;
}
/**
* 不斷產生[start,end]之間的隨機數，直到產生到target爲至
*/
void P01(int start, int end, int target) {
srand(time(NULL));
int temp, len = 0;
if (start > end)
swap(&start, &end);
if (target < start || target > end)
printf("%d is not in in the range of [%d,%d]\n!", target, start, end);
else {
do {
temp = doss(start, end);
printf("%d->", temp);
len++;
} while (temp != target);
printf("\nAfter %d times to get %d\n", len, temp);
}
}
/**
* 最大公約數
*/
int gcd(int a, int b) {
return a ? gcd(b % a, a) : b;
}
/**
* 最小公倍數
*/
int lcm(int a, int b) {
return a / gcd(a, b) * b;
}
/**
* 計算[start,end]之間被r整除的所有數之和，常數算法（等差數列的方法）
*/
int sum(int start, int end, int r) {
if (start > end)
swap(&start, &end);
start = (start + r - 1) / r * r;
end = end / r * r;
int n = (end - start) / r + 1;
return (start + end) * n >> 1;
}
void P02(int start, int end, int p, int q) {
if (start > end)
swap(&start, &end);
printf("%d\n", sum(start, end, lcm(p, q)));
}
/**
* 打印數字菱形
*/
void P03(int size) {
int i, j, n = 2 * size - 1;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++)
if (ABS(i-size) + ABS(j-size) >= size)//函數方法
putchar(' ');
else
putchar('A' - 1 + size - ABS(i-size) - ABS(j-size));
NEWLINE;
}
}
/**
* 不定方程x+y+z==total &&pa*x+pb*y+pc*z==money的優化枚舉通用方法
*/
void uniSolve(double pa, double pb, double pc, int total, int money) {
int a, b, c;
for (a = 0; a <= MIN(total,money/pa); a++)
for (b = 0; b <= MIN(total-a,money/pb); b++) {
c = total - a - b;
if (FEQUAL(a*pa+b*pb+c*pc,money))
printf("%d\t%d\t%d\n", a, b, c);
}
}
void P04() {
uniSolve(3, 2, 1 / 2., 100, 100);
}
void P05() {
uniSolve(2, 4, 2. / 9, 100, 100);
}
/**
* 趙錢孫李周的模擬法，用01234模擬圓桌的坐法
*/
int isLeft(int a, int b) {//這是基礎模型
return b == a + 1 || (a == 4 && b == 0);
}
int isRight(int a, int b) {//調用isLeft簡化定義
return isLeft(b, a);
}
int isNeighbour(int a, int b) {//同上
return isLeft(a, b) || isRight(a, b);
}
void P06() {
int s[] = { 1, 2, 3, 4 };
int zhao = 0, qian, sun, li, zhou;//因爲是環形，所以可以固定趙的座位
int zp, zq, lp, lq;
do {
shuffle(s, 4);//和下一行一起產生1,2,3,4的隨機排列賦給錢孫李周
qian = s[0], sun = s[1], li = s[2], zhou = s[3];
zp = isNeighbour(zhao, qian);
zq = (isLeft(sun, qian) || isLeft(li, qian));
lp = isLeft(qian, sun);
lq = isNeighbour(li, sun);
} while (zp || zq || lp || lq);//反覆找直到找到一組答案
printf("%d ", zhao);
printArr(s, 4);
}
/**
* 質數檢測的簡單優化算法
*/
int isPrime(int n) {
n = ABS(n);
if(n<2)return 0;
if(n==2)return 1;
if(n%2==0)return 0;
int t = sqrt(n), i;
for (i = 3; i <= t; i+=2)
if (n % i == 0)
return 0;
return 1;
}
void P07() {
int start = 100, end = 999, h, t, d, i, cnt = 0;
for (i = start; i <= end; i++) {
h = i / 100, t = (i / 10) % 10, d = i % 10;
if (h != t && t != d && d != h && t > h + d && !isPrime(t + h)) {
printf("%d ", i);
if (++cnt % 5 == 0)
NEWLINE;
}
}
NEWLINE;
}
//指派問題的窮舉法
void P08() {
int a, b, c, d, e;
for (a = 0; a <= 1; a++)
for (b = 0; b <= 1; b++)
for (c = 0; c <= 1; c++)
for (d = 0; d <= 1; d++)
for (e = 0; e <= 1; e++) {
if (a + b + c + d + e == 3 && a + c != 2 && b + c != 0
&& IMPLY(c,d+e==1) && b + c + d != 3
&&IMPLY(b,d+e!=2)) {
if (a)
putchar('A');
if (b)
putchar('B');
if (c)
putchar('C');
if (d)
putchar('D');
if (e)
putchar('E');
NEWLINE;
}
}
NEWLINE;
}
void P09() {
char line[80], buffer[80], *p = line;
int i = 0;
gets(line);
while (*p) {
i = 0;
while (isdigit(*p)) {
buffer[i++] = *p++;
}
buffer[i++] = '\0';
if (strlen(buffer))
puts(buffer);
buffer[0] = 0;
p++;
}
}
void P10() {
int s[] = { 1, 2, 3 };
int a, b, c, ap, aq, bp, bq, cp, cq;
do {
shuffle(s, 3);
a = s[0], b = s[1], c = s[2];
ap = (b == 2 && c == 2), aq = (a == 1);
bp = (b <= 2), bq = (c == 3);
cp = (a != 2), cq = (b != 1);
} while (!(a + ap + aq == 3 && b + bp + bq == 3 && c + cp + cq == 3));
printArr(s, 3);
}
//分糖果
void P11() {
int a, A, b, B, c, C, d, D, n = 64;
for (A = n / 2; A <= n; A++)
for (B = 0; B <= n - A; B++)
for (C = 0; C <= n - B - C; C++) {
D = n - A - B - C;
a = A, b = B, c = C, d = D;
a -= b + c + d, b += b, c += c, d += d;
b -= a + c + d, a += a, c += c, d += d;
c -= a + b + d, a += a, b += b, d += d;
d -= a + b + c, a += a, b += b, c += c;
if (a == b && b == c && c == d)
printf("%d %d %d %d\n", A, B, C, D);
}
}
void P12() {
LL x, len, wid, head, mid, tail, step;
scanf("%lld", &x);
len = log10(x) + 1;
wid = (len + 2) / 3;
step = powf(10, wid);
tail = x % step;
x /= step;
mid = x % step;
x /= step;
head = x;
printf("%lld\t%lld\t%lld\n", head, mid, tail);
printf("%lld", head + tail - mid);
}
void P13() {
char line[80], buffer[80], *p = line;
int i = 0;
gets(line);
while (*p) {
i = 0;
while (isalpha(*p)) {
buffer[i++] = *p++;
}
buffer[i++] = '\0';
if (strlen(buffer))
puts(buffer);
buffer[0] = 0;
p++;
}
}
//指派問題14
void P14() {
int a, b, c, d;
int ta, tb, tc, td;
for (a = 0; a <= 1; a++)
for (b = 0; b <= 1; b++)
for (c = 0; c <= 1; c++)
for (d = 0; d <= 1; d++) {
ta = d, tb = c, tc = !c, td = !tc;
if (ta + tb + tc + td == 2 && a + b + c + d == 1)
printf("%d %d %d %d\n", a, b, c, d);
}
}
//捕魚遞推方法
void P15(int n) {
int a[100], i;
a[n - 1] = 1;
for (i = n - 2; i >= 0; i--)
a[i] = n * a[i + 1] + 1;
for (i = 0; i < n; i++)
printf("%d ", a[i]);
NEWLINE;
}
//輔助函數：計算整數n的各位數字之和，也可以轉化爲字符串來求
int digitSum(int n) {
int s;
for (s = 0; n > 0;) {
s += n % 10;
n /= 10;
}
return s;
}
void P16(int len) {
int i, cnt = 4;
for (i = powf(10, len - 1); i < powf(10, len); i++) {
if (i % digitSum(i) == 0) {
printf("%d ", i);
if (++cnt % 5 == 0)
NEWLINE;
}
}
}
//一種直接的效率較低的搜索方法
void P17() {
int i, p, q;
for (i = 0;; i++) {
p = i * 10 + 6;
q = powf(10, ((int) log10(i) + 1)) * 6 + i;
if (q == 4 * p) {
WRITELN(p);
break;
}
}
}
/**
* 統計整數n的十進制表示中digit的個數digit是一位數
*/
int count(int n, int digit) {
int s;
for (s = 0; n > 0;) {
if (n % 10 == digit)
s++;
n /= 10;
}
return s;
}
//小明看書問題
void P18(int n) {
int i, sum = 0;
for (i = 1; sum < n; i++)
sum += count(i, 1);
if (sum == n)
WRITELN(i);
else //這個容易忽視
puts("IMPOSSIBLE!");
}
void P19() {
int a, b;
READ(a);
READ(b);
printf("%6d\n+%5d\n--------\n%6d\n", a, b, a + b);
}
void P20() {
uniSolve(3, 2, 1, 30, 50);
}
int main(int argc, char **argv) {
srand(time(NULL));
return EXIT_SUCCESS;
}