FZU 2156 Climb Stairs

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题目大意：你要从0层开始，经过A，和B层，到达N层，每次可以走x或者y层，输出方法数。

比较简单的dp了吧。。只是处理方式稍微机智点（虽然还是比较好想到的），递推方程dp[i] = dp[i - x] + dp[i - y],然后输出dp[A] + dp[B - A] + dp[N - B]，（事先swap保证A <= B) 。

#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <string>#include <cctype>#include <cstdio>#include <vector>#include <cstdlib>#include <cstring>#include <iomanip>#include <sstream>#include <iostream>#include <algorithm>using namespace std;#define ls id<<1,l,mid#define rs id<<1|1,mid+1,r#define OFF(x) memset(x,-1,sizeof x)#define CLR(x) memset(x,0,sizeof x)#define MEM(x) memset(x,0x3f,sizeof x)typedef long long ll ;typedef pair<int,int> pii ;const int maxn = 1e5+50 ;const int inf = 0x3f3f3f3f ;const int MOD = 1e9+7 ;ll dp[maxn];int n, x, y, a, b;int main () {#ifdef LOCALfreopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);//      freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif    while (cin >> n >> x >> y >> a >> b) {        CLR(dp); dp[0] = 1;        for (int i = 1; i <= n; i++) {            if (i >= x) dp[i] = (dp[i] + dp[i - x]) % MOD;            if (i >= y) dp[i] = (dp[i] + dp[i - y]) % MOD;        }        if (a > b) swap(a, b);        printf("%I64d\n", (dp[a] * dp[b - a]) % MOD * dp[n - b] % MOD);    }return 0;}

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