SRM 605 D1 L2：AlienAndSetDiv1，DP，bitmask

 

跟D2 L3 類似，唯一的區別是D2L3是兩數之差至多為K, 這題是至少K，增加了一點難度，DP的狀態不好想。類似D2L3，只考慮未匹配的數，不過將未匹配的數分成了兩類，一類是Good integers，g >= n + K，另一類是, s > n && s < n + K。以從大到小的順序考慮2N個數。

代碼如下：

 

#include #include #include #include #include #include #include #include #include #include#include #include #include #include #include using namespace std;/*************** Program Begin **********************/const int MOD = 1000000007;  const int MAX_N = 50;const int MAX_K = 10;int dp[2 * MAX_N + 1][2 * MAX_N + 1][1 << MAX_K];class AlienAndSetDiv1 {public:int K;int calc(int s, int g, int n){int res = 0;if (-1 != dp[n][g][s]) {return dp[n][g][s];}if (n == 0) {if (0 == s && 0 == g) {// base caseres = 1;}} else {if (0 == s && 0 == g) {if (1 == K) {res += 2 * calc(0, 1, n-1);// add} else {res += 2 * calc(1, 0, n-1);// add}res %= MOD;} else if (0 == s && g != 0) {res += calc(0, g-1, n-1);// matchif (1 == K) {res += calc(0, g+1, n-1);// add} else {res += calc(1, g, n-1);// add}res %= MOD;} else if (s != 0 && 0 == g) {// K != 1，只能addint newset = s;int i = 0;  for (i = 0; (newset & 0x80000000 ) == 0; newset = ( newset << 1 ), ++i) {  // 注意位操作的優先順序，要加括弧}  int mx = 31 - i;    // mx 為s不為0的最高位if (n + mx + 1 - (n - 1) >= K) {// g -> 1int t = (s - (1 << mx));// 清除mx位res += calc( (t << 1) | 1, 1, n - 1 );// addres %= MOD;} else {res += calc( (s << 1) | 1, 0, n - 1 );// addres %= MOD;}} else {// s != 0 && 0 != g// K != 1int newset = s;int i = 0;  for (i = 0; (newset & 0x80000000 ) == 0; newset = ( newset << 1 ), ++i) {}  int mx = 31 - i;    // mx 為s不為0的最高位if ( n + mx + 1 - (n - 1) >= K) {// 判斷是否使g變化int t = (s - (1 << mx));res += calc(t << 1, g, n-1);// matchres += calc( (t << 1) | 1, g+1, n - 1 ); // add} else {res += calc(s << 1, g-1, n-1);// matchres += calc( (s << 1) | 1, g, n - 1 );// add}res %= MOD;}}dp[n][g][s] = res;return res;}int getNumber(int N, int K) {int res = 0;this->K = K;memset(dp, -1, sizeof(dp));res = calc(0, 0, 2 * N);return res;}};/************** Program End ************************/