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http://acm.hdu.edu.cn/showproblem.php?pid=5640
这题有点辗转相除的意思。基本没有什么坑点。
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <vector> #include <string> #define LL long long using namespace std; int n, m; int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { scanf("%d%d", &n, &m); int ans = 0; while (n&&m) { if (n > m) swap(n, m); ans += m/n; m = m%n; } printf("%d\n", ans); } return 0; }
http://acm.hdu.edu.cn/showproblem.php?pid=5641
这题坑点有点多，3A。。首先任意一条连线，需要判断中间有没有经过什么点。
然后不能有重复的点，需要判断。
其次，对于0和大于9的数据需要判断掉。
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <vector> #include <string> #define LL long long using namespace std; int k, s[10]; bool vis[10]; int cross(int from, int to) { if (from > to) swap(from, to); if (from == 1 && to == 3) return 2; if (from == 1 && to == 7) return 4; if (from == 1 && to == 9) return 5; if (from == 2 && to == 8) return 5; if (from == 3 && to == 7) return 5; if (from == 4 && to == 6) return 5; if (from == 3 && to == 9) return 6; if (from == 7 && to == 9) return 8; return 0; } bool work() { if (k < 4) return false; memset(vis, false, sizeof(vis)); int t; for (int i = 1; i < k; ++i) { if (s[i-1] == 0 || s[i] == 0) return false; if (s[i-1] > 9 || s[i] > 9) return false; if (s[i] == s[i-1]) return false; if (vis[s[i]]) return false; t = cross(s[i-1], s[i]); if (t && !vis[t]) return false; vis[s[i-1]] = true; vis[s[i]] = true; } return true; } int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { scanf("%d", &k); for (int i = 0; i < k; ++i) scanf("%d", &s[i]); if (work()) printf("valid\n"); else printf("invalid\n"); } return 0; }
http://acm.hdu.edu.cn/showproblem.php?pid=5642
这题是个递推，但是没考虑清楚，递推式一直是错的，4A。。
设p[n][k]表示n长度字符串，结尾k个相同的情况数。
那么：
p[i][1] = 25*(p[i-1][1]+p[i-1][2]+p[i-1][3])%MOD;
p[i][2] = p[i-1][1];
p[i][3] = p[i-1][2];
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <vector> #include <string> #define LL long long #define MOD 1000000007 using namespace std; const int maxN = 2005; int n; LL p[maxN][5]; void init() { p[1][1] = 26; p[1][2] = 0; p[1][3] = 0; for (int i = 2; i < maxN; ++i) { p[i][1] = 25*(p[i-1][1]+p[i-1][2]+p[i-1][3])%MOD; p[i][2] = p[i-1][1]; p[i][3] = p[i-1][2]; } } int main() { //freopen("test.in", "r", stdin); init(); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { scanf("%d", &n); cout << (p[n][1]+p[n][2]+p[n][3])%MOD << endl; } return 0; }
http://acm.hdu.edu.cn/showproblem.php?pid=5643
这一题想法类似于O(n)做法的约瑟夫问题。需要考虑子问题，然后映射回来。具体的方法百度百科里有。
最后
p(n, k) = (p(n-1, k+1)+k)%n or n(if 0)
不过我二逼的离线了一发，然后MLE了。。
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <vector> #include <string> #define LL long long using namespace std; const int maxN = 5001; /* int p[maxN][maxN]; void init() { for (int i = 0; i < maxN; ++i) p[1][i] = 1; for (int i = 2; i < maxN; ++i) { for (int j = 1; j < maxN; ++j) { p[i][j] = (p[i-1][j+1]+j)%i; if (!p[i][j]) p[i][j] = i; } } } */ int dfs(int n, int k) { if (n == 1) return 1; int ans; ans = (dfs(n-1, k+1)+k)%n; if (!ans) ans = n; return ans; } int main() { //freopen("test.in", "r", stdin); //init(); int n, T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { scanf("%d", &n); printf("%d\n",dfs(n, 1)); } return 0; }
1005:没有想法。。应该是水平还没到火候。。。
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原文地址：http://www.cnblogs.com/andyqsmart/p/5270278.html
