剑指10-II动态规划：青蛙跳台阶 搜索和回溯算法 求1+2+...n 天秀

class Solution {
    #define mod 1000000007
public:
    int numWays(int n) {
        int result;
        if(n==0) return 1;
        int x=0,y=0,z=1;
        for(int i=0;i<n;i++)
        {
            x=y;
            y=z;
            z=(x+y)%mod;
        }
        return z;
    }
};

class Solution {
    #define mod 1000000007
public:
    int numWays(int n) {
        int f[101];
        f[0]=f[1]=1;
        for(int i=2;i<=n;i++)
        {
            f[i]=(f[i-1]+f[i-2])%mod;
        }
        return f[n];
    }
};

class Solution {
public:
    int sumNums(int n) {
        bool a[n][n+1];
        return sizeof(a)>>1;
    }
};