POJ 1953

题意：求出长度为n的01串中，不出现连续1的串的个数

思路：

dp0[x]表示以0开头长度为x的这类01串的个数，dp1[x]表示以1开头长度为x的这类01串的个数

dp0[x] = dp0[ x-1 ] + dp1[ x-1 ], dp0[1] = 1;

dp1[x] = dp0[ x-1 ] , dp1[1] = 1;

 

总结：要用long long，开始错了两次，应该先测试最大数据再提交的，囧了。0ms

 

#include <iostream>#define F(i,a,b) for (int i=a;i<=b;i++)using namespace std;long long dp0[50], dp1[50];long long dfs1(int x);long long dfs0(int x){if (x == 1 )return 1;if ( dp0[x] )return dp0[x] ;dp0[x] = dfs0(x-1) + dfs1(x-1);return dp0[x];}long long dfs1( int x ){if (x == 1) return 1;if ( dp1[x] )return dp1[x];dp1[x] = dfs0(x-1);return dp1[x];}int main(){int n, T;cin >> T;memset(dp0, 0, sizeof(dp0));memset(dp1, 0, sizeof(dp1));F(t,1,T){cin >> n;cout << "Scenario #" << t << ":/n";cout << dfs0(n) + dfs1(n) << endl << endl;}return 0;}