中國大學MOOC-PKU-程序設計與算法（二）算法基礎-題解

一. 枚舉

1. 2810:完美立方

http://bailian.openjudge.cn/practice/2810/
Solution：四重循環。

#include <iostream>

using namespace std;
int main() {
	int N;
	cin >> N;
	for(int a = 2; a <= N; ++a) {
		for(int b = 2; b <= a - 1; ++b) {
			for(int c = b; c <= a - 1; ++c) {
				for(int d = c; d <= a - 1; ++d) {
					if(a * a * a == b * b * b + c * c * c + d * d * d) {
						printf("Cube = %d, Triple = (%d,%d,%d)\n", a, b, c, d);
					}
				}
			}
		}
	}
	return 0;
}

2. 4148:生理週期

http://bailian.openjudge.cn/practice/4148/
Solution：所求天數(k-d)<=21252，說明天數k<=21252+d。

#include <iostream>
#define N 21252

using namespace std;
int main() {
	int p, e, i, d, caseNo = 1;
	while(cin >> p >> e >> i >> d && p != -1) {
	    for(int k = d + 1; k <= N + d; ++k) {
	        if((k - p) % 23 == 0 && (k - e) % 28 == 0 && (k - i) % 33 == 0) {
                printf("Case %d: the next triple peak occurs in %d days.\n", caseNo++, k - d);
                break;
	        }
	    }
	}
	return 0;
}

更快地完成for循環：

#include <iostream>
#define N 21252

using namespace std;
int main() {
	int p, e, i, d, caseNo = 1;
	while(cin >> p >> e >> i >> d && p != -1) {
        int k;
        for(k = d + 1; (k - p) % 23; ++k);
        for(; (k - e) % 28; k += 23);
        for(; (k - i) % 33; k += 23 * 28);
        printf("Case %d: the next triple peak occurs in %d days.\n", caseNo++, k - d);
	}
	return 0;
}

3. 2692:假幣問題

http://bailian.openjudge.cn/practice/2692/
Solution：枚舉每一個硬幣爲假且重或輕的情況。
<cstring> strchr：http://www.cplusplus.com/reference/cstring/strchr/

#include <iostream>
#include <cstring>

using namespace std;

char Left[3][7];
char Right[3][7];
char result[3][7];
bool isFake(char c, bool isLight);

int main() {
	int n;
	cin >> n;
	while(n--) {
	    for(int i = 0; i < 3; ++i) {
	        cin >> Left[i] >> Right[i] >> result[i];
	    }
	    for(char c = 'A'; c <= 'L'; ++c) {
	        if(isFake(c, true)) {
	            cout << c << " is the counterfeit coin and it is light. " << endl;
	            break;
	        }
	        else if(isFake(c, false)) {
	            cout << c << " is the counterfeit coin and it is heavy. " << endl;
	            break;
	        }
	    }
	}
	return 0;
}

bool isFake(char c, bool isLight) {
    for(int i = 0; i < 3; ++i) {
        char* pLeft, *pRight;
        if(isLight) {
            pLeft = Left[i];
            pRight = Right[i];
        }
        else {
            pLeft = Right[i];
            pRight = Left[i];
        }
        switch(result[i][0]) {
            case 'u':
                if (strchr(pRight, c) == NULL) {
                    return false;
                }
                break;
            case 'e':
                if (strchr(pLeft, c) || strchr(pRight, c)) {
                    return false;
                }
                break;
            case 'd':
                if (strchr(pLeft, c) == NULL) {
                    return false;
                }
                break;
        }
    }
    return true;
}

二. 遞歸（一）

1.