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

Chocolate in its many forms is enjoyed by millions of people around the world every day. It is a truly universal candy available in virtually every country around the world.

You find that the only thing better than eating chocolate is to share it with friends. Unfortunately your friends are very picky and have different appetites: some would like more and others less of the chocolate that you offer them. You have found it increasingly difficult to determine whether their demands can be met. It is time to writte a program that solves the problem once and for all!

Your chocolate comes as a rectangular bar. The bar consists of same-sized rectangular pieces. To share the chocolate you may break one bar into two pieces along a division between rows or columns of the bar. You or the may then repeatedly break the resulting pieces in the same manner. Each of your friends insists on a getting a single rectangular portion of the chocolate that has a specified number of pieces. You are a little bit insistent as well: you will break up your bar only if all of it can be distributed to your friends, with none left over.

For exampla, Figure 9 shows one way that a chocolate bar consisting of 3 x 4 pieces can be split into 4 parts that contain 6, 3, 2, and 1 pieces respectively, by breanking it 3 times (This corresponds to the first sample input.)

Input 

The input consists of multiple test cases each describing a chocolate bar to share. Each description starts with a line containing a single integer n (1n15), the number of parts in which the bar is supposed to be split. This is followed by a line containing two integers x and y (1x, y100), the dimensions of the chocolate bar. The next line contains n positive integers, giving the number of pieces that are supposed to be in each of the n parts.

The input is terminated by a line containing the integer zero.

Output 

For each test case, first display its case number. Then display whether it is possible to break the chocolate in the desired way: display ``Yes" if it is possible, and ``No" otherwise. Follow the format of the sample output.

Sample Input 

4 
3 4 
6 3 2 1 
2 
2 3 
1 5 
0

Sample Output 

Case 1: Yes 
Case 2: No

---

做法 : DP+bitmask
加了一堆剪枝才過, 此題非同小可 !

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h> 
#define Done 1
typedef struct Node {
	int length, width;
	int state, next;
};
Node node[32768*50];
int size;
int A[20], n, x, y, i;
int Link[32768], Sqrt[10001];
int LOG2[32768], Sum[32768];
char Used[101], mark[32768][101], times;
int check(int length, int width) {
	if(length <= x && width <= y)
		return 1;
	if(length <= y && width <= x)
		return 1;
	return 0;
}
void insertList(int state, int length, int width) {
	if(check(length, width) == 0)	return;
	if(mark[state][length] != 0)	return;
	static int prev, curr;
	prev = 0, curr = Link[state];
	while(curr != 0) {
		if(node[curr].length < length)	break;
		else if(node[curr].length == length)
			return;
		else
			prev = curr, curr = node[curr].next;
	}
	size++;
	if(prev == 0)	Link[state] = size;
	else			node[prev].next = size;
	node[size].length = length;
	node[size].width = width;
	node[size].state = state;
	node[size].next = curr;
	mark[state][length] = width;
	mark[state][width] = length;
	if(Used[length])
		Used[length]--, times--;
	if(Used[width])
		Used[width]--, times--;
}
void doCombine(int state1, int state2) {
	static int i, j, state;
	static int a, b, c, d;
	state = state1 | state2;
	for(i = Link[state1]; i; i = node[i].next) {
		a = node[i].length, b = node[i].width;
		if(mark[state2][a] != 0) {
			insertList(state, a, b+mark[state2][a]);
			insertList(state, b+mark[state2][a], a);
		}
		if(mark[state2][b] != 0) {
			insertList(state, b, a+mark[state2][b]);
			insertList(state, a+mark[state2][b], b);
		}
		if(times == 0)	return;
	}
}
void doBuild() {
	int i, maxState, minState;
	int state, state1, state2;
	int s, finalState, sqr, tmp;
	finalState = (1<<n)-1;
	for(state = 1; state <= finalState; state++) {
		s = LOG2[state&(-state)];
		maxState = (state>>s)-1;
		minState = 1;
		Sum[state] = Sum[state-(state&(-state))]+Sum[state&(-state)];
		sqr = Sqrt[Sum[state]], tmp = Sum[state], times = 0;
		memset(Used, 0, sizeof(Used));
		for(i = 1; i <= sqr; i++)
			if(tmp%i == 0 && check(i, tmp/i) == 1) {
				Used[i]++, Used[tmp/i]++;
				times += 2;
			}
		for(i = minState; i <= maxState; i++) {
			if((state&(i<<s)) == (i<<s)) {
				state1 = i<<s;
				state2 = state ^ state1;
				if(state1 > state2)	break;
				doCombine(state1, state2);
				if(times == 0)		break;
			}
		}
	}
	if(mark[finalState][x] != 0)
		puts("Yes");
	else
		puts("No");
}
int initBuild() {
	int i, j, sqr, finalState;
	size = 0, finalState = (1<<n)-1;
	for(i = 0; i <= finalState; i++)
		Link[i] = 0, memset(mark[i], 0, sizeof(mark[i]));
	for(i = 0; i < n; i++) {
		sqr = Sqrt[A[i]];
		for(j = 1; j <= sqr; j++) {
			if(A[i]%j == 0 && check(j, A[i]/j) == 1) {
				insertList(1<<i, j, A[i]/j);
				insertList(1<<i, A[i]/j, j);
			}
		}
		Sum[1<<i] = A[i];
	}
} 
int main() {
	int Case = 0, tmp, i;
	for(i = 0; i <= 32767; i++)
		LOG2[i] = (int)log2(i);
	for(i = 0; i <= 10000; i++)
		Sqrt[i] = (int)sqrt(i);
	while(scanf("%d", &n) == 1 && n) {
		scanf("%d %d", &x, &y);
		int sum = 0;
		for(i = 0; i < n; i++)
			scanf("%d", &A[i]), sum += A[i];
		printf("Case %d: ", ++Case);
		if(sum != x*y) {
			puts("No");
			continue;
		}
		initBuild();
		doBuild();
		
	}
    return 0;
}