題目戳這里。
找規律。

• 每一列占據的格子一定是一段區間；
• 相鄰列之間的區間有交。
• 上界先增后減，下界先減后增。

\(f_{i,j,k,0/1,0/1}\)表示考慮前\(i\)列，第\(i\)列，上界為\(j\)下界為\(k\)且上界正在上升/下降，下界正在上升/下降的方案數。轉移請自行YY。

#include<string> #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std;const int maxn = 60,ten[4] = {1,10,100,1000},maxl = 12; int N; struct BigNumber {int d[maxl];inline BigNumber(const string &s){memset(d,0,sizeof d);int len = s.size(),i,j,k; d[0] = (len-1)/4+1;for (i = 1;i < maxl;++i) d[i] = 0;for (i = len-1;i >= 0;--i){j = (len-i-1)/4+1,k = (len-i-1)%4;d[j] += ten[k]*(s[i]-'0');}while (d[0] > 1&& !d[d[0]]) --d[0];}inline BigNumber() { *this = BigNumber(string("0")); }inline string toString() const{string s(""); int i,j,temp;for (i = 3;i >= 1;--i) if (d[d[0]] >= ten[i]) break;temp = d[d[0]];for (j = i;j >= 0;--j) s += (char)(temp/ten[j]+'0'),temp %= ten[j];for (i = d[0]-1;i;--i){temp = d[i];for (j = 3;j >= 0;--j) s += (char)(temp/ten[j]+'0'),temp %= ten[j];}return s;}friend inline BigNumber operator +(const BigNumber &a,const BigNumber &b){BigNumber c; int x = 0; c.d[0] = max(a.d[0],b.d[0]);for (int i = 1;i <= c.d[0];++i) x += a.d[i]+b.d[i],c.d[i] = x % 10000,x /= 10000;while (x) c.d[++c.d[0]] = x % 10000,x /= 10000;return c;}friend inline bool operator ==(const BigNumber &a,const BigNumber &b){if (a.d[0] != b.d[0]) return false;for (int i = 1;i <= a.d[0];++i) if (a.d[i] != b.d[i]) return false;return true;} }f[maxn][maxn][maxn][2][2],ans[maxn*2];inline void ready() {for (int r = 1;r <= 25;++r){if (r == 2){int u; ++u;}int c = 50-r;for (int i = 1;i <= r;++i)for (int j = i;j <= r;++j){f[1][i][j][0][0] = BigNumber(string("1"));f[1][i][j][0][1] = f[1][i][j][1][0] = f[1][i][j][1][1] = BigNumber();if (i == 1) f[1][i][j][1][0] = BigNumber(string("1"));if (j == r) f[1][i][j][0][1] = BigNumber(string("1"));if (i == 1&&j == r) f[1][i][j][1][1] = BigNumber(string("1"));}for (int i = 2;i <= c;++i)for (int j = 1;j <= r;++j)for (int k = j;k <= r;++k){f[i][j][k][0][0] = f[i][j][k][0][1] = f[i][j][k][1][0] = f[i][j][k][1][1] = BigNumber();for (int jj = 1;jj <= r;++jj)for (int kk = jj;kk <= r;++kk){if (jj > k||kk < j) continue;if (jj >= j){if (kk <= k) f[i][j][k][0][0] = f[i-1][jj][kk][0][0]+f[i][j][k][0][0];if (kk >= k&&k != r) f[i][j][k][0][1] = f[i-1][jj][kk][0][1]+f[i][j][k][0][1]; }if (jj <= j&&j != 1){if (kk <= k) f[i][j][k][1][0] = f[i][j][k][1][0]+f[i-1][jj][kk][1][0];if (kk >= k&&k != r) f[i][j][k][1][1] = f[i][j][k][1][1]+f[i-1][jj][kk][1][1];}}if (j == 1&&k == r){f[i][j][k][1][1] = f[i][j][k][0][0];f[i][j][k][0][1] = f[i][j][k][0][0];f[i][j][k][1][0] = f[i][j][k][0][0];}else if (j == 1) f[i][j][k][1][0] = f[i][j][k][0][0],f[i][j][k][1][1] = f[i][j][k][0][1];else if (k == r) f[i][j][k][0][1] = f[i][j][k][0][0],f[i][j][k][1][1] = f[i][j][k][1][0];}for (c = r;c <= 49;++c){if (r+c > 50) continue;for (int i = 1;i <= r;++i)for (int j = i;j <= r;++j){ans[(r+c)<<1] = ans[(r+c)<<1]+f[c][i][j][1][1];if (c != r) ans[(r+c)<<1] = ans[(r+c)<<1]+f[c][i][j][1][1];}}} }int main() {freopen("10884.in","r",stdin);freopen("table.out","w",stdout);ready();printf("ans[101]={");for (int i = 0;i <= 100;++i){if (i) putchar(',');cout << "string(\""<<ans[i].toString() << "\")";}putchar('}');fclose(stdin); fclose(stdout);return 0; }

由于常數寫丑了，用上面的程序打了個表。

#include<string> #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std;int N,T; string ans[101]={string("0"),string("0"),string("0"),string("0"),string("1"),string("0"),string("2"),string("0"),string("7"),string("0"),string("28"),string("0"),string("120"),string("0"),string("528"),string("0"),string("2344"),string("0"),string("10416"),string("0"),string("46160"),string("0"),string("203680"),string("0"),string("894312"),string("0"),string("3907056"),string("0"),string("16986352"),string("0"),string("73512288"),string("0"),string("316786960"),string("0"),string("1359763168"),string("0"),string("5815457184"),string("0"),string("24788842304"),string("0"),string("105340982248"),string("0"),string("446389242480"),string("0"),string("1886695382192"),string("0"),string("7955156287456"),string("0"),string("33468262290096"),string("0"),string("140516110684832"),string("0"),string("588832418973280"),string("0"),string("2463133441338048"),string("0"),string("10286493304041104"),string("0"),string("42892130604098656"),string("0"),string("178592047539343200"),string("0"),string("742609229473744320"),string("0"),string("3083957343567791392"),string("0"),string("12792021060576424896"),string("0"),string("53000868925259947840"),string("0"),string("219365134324873522816"),string("0"),string("907023528883142832360"),string("0"),string("3746790354386182679408"),string("0"),string("15463645062002474062384"),string("0"),string("63767018378178067474656"),string("0"),string("262742756317344213209200"),string("0"),string("1081765434874991509707040"),string("0"),string("4450606984357021640248032"),string("0"),string("18298022787758605020282816"),string("0"),string("75179913955330333724697136"),string("0"),string("308691924054843201409922592"),string("0"),string("1266737680502193374869298720"),string("0"),string("5195143014579351011947302208"),string("0"),string("21294548056433354780482923744"),string("0"),string("87238762619153966026251258944"),string("0"),string("357215388993130669706869321408")};int main() {freopen("10884.in","r",stdin);freopen("10884.out","w",stdout);scanf("%d",&T);for (int Case = 1;Case <= T;++Case){printf("Case #%d: ",Case); scanf("%d",&N);cout << ans[N] << endl;}fclose(stdin); fclose(stdout);return 0; }

轉載于:https://www.cnblogs.com/mmlz/p/6403879.html

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