2017"百度之星"程序设计大赛 - 初赛（A）

1001

统计p - 1的因子个数

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;

int main(){
	int t;	scanf("%d", &t);
	while(t--) {
		int n;	scanf("%d", &n);
		n = n - 1;
		int ans = 1;
		for(int i = 2; i * i <= n; ++i) {
			if(n % i == 0) {
				int cnt = 1;
				while(n % i == 0) {
					n /= i;
					cnt++;
				}
				ans *= cnt;
			}
		}
		if(n > 1)	ans *= 2;
		printf("%d\n", ans);
	}
	return 0;
}

1005

暴力计算即可，要注意的就是2月29以及2月前后的分析

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;

int main(){
	int	t;	scanf("%d", &t);
	while(t--) {
		int d, m, y;
		scanf("%d-%d-%d", &y, &m, &d);
		int cnt = 0;
		if(m == 2 && d == 29) {
			for(int i = y + 1; ; ++i) {
				if((i % 4 == 0 && i % 100 != 0) || (i % 400 == 0)) {
					cnt += 2;
					if(cnt > 2 && cnt % 7 == 0) {
						printf("%d\n", i);
						break;
					}
				} else {
					cnt += 1;
				}
			} 
		} else if(m <= 2) {
			for(int i = y; ; ++i) {
				if((i % 4 == 0 && i % 100 != 0) || (i % 400 == 0)) {
					cnt += 2;
				} else {
					cnt += 1;
				}
				if(cnt > 0 && cnt % 7 == 0) {
					printf("%d\n", i + 1);
					break;
				}
			}
		} else if(m > 2) {
			for(int i = y + 1 ; ; ++i) {
				if((i % 4 == 0 && i % 100 != 0) || (i % 400 == 0)) {
					cnt += 2;
				} else {
					cnt += 1;
				}
				if(cnt > 0 && cnt % 7 == 0) {
					printf("%d\n", i);
					break;
				}
			}
		}
	}
	return 0;
}

1006

直接数1和0的联通块数就好了，但在此之前要先在外围涂上一层0

刘汝佳的紫书上有类似的题 160+页吧、

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 105 + 10;
const int INF = 1e9 + 10;
char st[qq][qq];
int vis[qq][qq];
int n, m;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
void Dfs(int x, int y, char ch) {
	vis[x][y] = 1;
	for(int i = 0; i < 4; ++i) {
		int a = x + dx[i];
		int b = y + dy[i];
		if(a < 0 || b < 0 || a > n + 1 || b > m + 1)	continue;
		if(st[a][b] != ch)	continue;
		if(vis[a][b])	continue;
		Dfs(a, b, ch);
	}
}

int main(){
	while(scanf("%d%d", &n, &m) != EOF) {
		for(int i = 1; i <= n; ++i) {
			scanf("%s", st[i] + 1);
		}
		for(int i = 0; i <= m + 1; ++i) {
			st[0][i] = st[n + 1][i] = '0';
		}
		for(int i = 0; i <= n + 1; ++i) {
			st[i][0] = st[i][m + 1] = '0';
		}
		mst(vis, 0);
		int cnt0, cnt1;
		cnt0 = cnt1 = 0;
		for(int i = 0; i <= n + 1; ++i) {
			for(int j = 0; j <= m + 1; ++j) {
				if(vis[i][j])	continue;
				Dfs(i, j, st[i][j]);
				if(st[i][j] == '0')	cnt0++;
				else	cnt1++;
			}
		}
		if(cnt1 != 1) {
			puts("-1");
		} else {
			if(cnt0 == 1)	puts("1");
			else if(cnt0 == 2)	puts("0");
			else	puts("-1");
		}
	}
	return 0;
}