不慌不忙、不急不躁

LightOJ 1224 - DNA Prefix题目大意：给定若干个字符串，找到一个前缀，前缀长度出现次数最大值。解题思路：对字符串集合建立字典树，然后遍历一遍，每个节点等于depval。#include #include #include using namespace std;const int maxn = 50005 * 50;const int sigma

题目链接：LightOJ 1044 - Palindrome Partitioning代码#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;const int maxn = 1005;const int inf = 0x3f

题目链接：LightOJ 1018 - Brush (IV)代码#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int maxn = 17;const int maxs = 1<<17;const int inf = 0x3f3f3f3f;in

题目链接：LightOJ 1025 - The Specials Menu代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 65;typedef long long ll;char s[maxn];ll dp[maxn][maxn];ll solve (

题目链接：LightOJ 1047 - Neighbor House代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 30;const int inf = 0x3f3f3f3f;int N, dp[maxn][3];int main () { in

题目链接：LightOJ 1060 - nth Permutation代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 30;typedef long long ll;int C[maxn][maxn];char S[maxn];int N, K, c

题目链接：LightOJ 1050 - Marbles代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 505;int R, B;double dp[maxn][maxn];void init () { memset(dp, 0, sizeof(d

题目链接：LightOJ 1004 - Monkey Banana Problem代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 205;int N, dp[maxn][maxn];int solve () { for (int i = 2; i

题目链接：LightOJ 1027 - A Dangerous Maze代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }int main () { int cas;

题目链接：LightOJ 1037 - Agent 47代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = (1<<15) + 5;const int maxm = 20;const int inf = 0x3f3f3f3f;int N, H[maxm],

题目链接：LightOJ 1021 - Painful Bases代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 25;const int maxs = 1<<17;typedef long long ll;int base, K;char s[ma

题目链接：LightOJ 1036 - A Refining Company代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 505;int N, M, dp[maxn][maxn];int A[maxn][maxn], B[maxn][maxn];//

题目链接：LightOJ 1033 - Generating Palindromes代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;int N, dp[maxn][maxn];char s[maxn], t[maxn];int solve ()

题目链接：LightOJ 1031 - Easy Game代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;int N, A[maxn], S[maxn], dp[maxn][maxn];bool vis[maxn][maxn];void init

题目链接：LightOJ 1129 - Consistency Checker题目大意：给定若干个字符串，判断是否存在前缀。解题思路：字典树的简单应用。#include #include #include using namespace std;const int maxn = 10005;const int sigma_size = 10;struct Ti

题目链接：LightOJ 1269 - Consecutive Sum题目大意：给定一个序列，选定一段区间的亦或和，输出最大和最小。解题思路：最大很简单，对所有前缀建立字典树，然后尽量往反向走；最小则需要往正向走，并且向正向走的时候要扣除自己本身。#include #include #include using namespace std;const int

题目链接：Light OJ 1348 - Aladdin and the Return Journey题目大意：给定一棵树，两种操作0 i j：ij路径上的权值和1 i v：将第i个节点的权值修改为v解题思路：树链剖分的裸题。#include #include #include using namespace std;const int maxn = 300

题目链接：LightOJ 1017 - Brush (III)代码#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 105;int N, W, M, G[maxn], Y[maxn];void init () { sc

题目链接：LightOJ 1030 - Discovering Gold代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;int N;double dp[maxn], A[maxn];int main () { int cas; s

题目链接：LightOJ 1032 - Fast Bit Calculations代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 50;ll dp[maxn + 5][maxn + 5][2];int main

题目链接：LightOJ 1057 - Collecting Gold代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 55;const int maxs = 1<<17;const int inf = 0x3f3f3f3f;int N, M, K, X

题目链接：LightOJ 1038 - Race to 1 Again代码#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 1e5;double dp[maxn + 5];vector<int> G[maxn + 5];vo

题目链接：LightOJ 1005 - Rooks代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;int N, K;ll C(int n, int k) { if (n - k < k) k = n - k; ll ret = 1;

题目链接：LightOJ 1011 - Marriage Ceremonies代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 20;const int maxm = 1<<17;int N, dp[maxn][maxm], W[maxn][maxn];i

题目链接：LightOJ 1013 - Love Calculator代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 50;typedef long long ll;int N, M, d[maxn][maxn];char a[maxn], b[max

题目链接：LightOJ 1051 - Good or Bad代码#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 55;int N;char S[maxn];bool dp[maxn][10][10];bool isvowel(char ch) {