随笔分类 -  dp

摘要：hdu 4563 + 把每个命令走的距离抽象成完全背包 + 枚举最后一个不是整点走完的命令 include include include include include include include using namespace std; typedef long long LL; const 阅读全文

摘要：```cpp include include include include include include include include define ll long long using namespace std; const int MOD = 1e8+7; int dp[55][1010 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #define ll long long 10 11 using namespace std; 12 13 int color[105],n; 14 int dp[105][105... 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #define ll long long 7 8 using namespace std; 9 const int N = 1e5+1000; 10 11 int dp[105*2][105][105]; 12 int maze[105][105]; ... 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #define ll long long 7 8 using namespace std; 9 const int N = 1e5+1000; 10 11 int a[N],dp[N]; 12 13 void solve() 14 { 15 ... 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #define ll long long 7 8 using namespace std; 9 const int N = 1e4+1000; 10 11 double dp[N]; 12 13 void solve() 14 { 15 in... 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 //#include 9 //#include 10 //#include 11 #define LL long long 12 13 using namespace std; 14 //using ... 阅读全文

摘要：1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 //#include 9 //#include 10 //#include 11 #define LL long long 12 13 using namespace std; 14 //using ... 阅读全文

摘要：面对位运算，一直很无感。。。可能数学太差，脑洞太小。 1.首先是最基本的： 与&，或|，非~，异或^。 2.获取一个或者多个固定位的值： 假设 x = 1010（二进制），我们要取左数第二位的值，可以用（x &（1<<1)）; 还可用（x&(3<<2))来取得第三位和第四位。 3.把一个或者多个固定 阅读全文

摘要：Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke turned into a student and read a book.Suddenly, a difficul... 阅读全文

摘要：Anniversary partyTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6990Accepted Submission(s): 3104P... 阅读全文

摘要：10036 - Divisibility额..直接复制不过来，只好叙述一下了...t组样例，n个数（1～10000），k（2～100）是要取余的数，然后给出n个数第一个数前不能加正负号，其他的数前面可以加正负号，然后问这些正负号任意加，所有的情况中是否有能对k取余得0的情况。若有输出：Di...否则... 阅读全文

摘要：DividingTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20635Accepted Submission(s): 5813Problem D... 阅读全文

摘要：Good NumbersTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3467Accepted Submission(s): 1099Proble... 阅读全文

摘要：1 #include 2 #include 3 #include ; 4 using namespace std; 5 #define MAX_N 5 6 #define INF 9999 7 int n=5; 8 int d[MAX_N][MAX_N]={ 9 {INF,3,... 阅读全文

摘要：龟兔赛跑Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13497Accepted Submission(s): 5050Problem Descr... 阅读全文