How do you add?(UVA 10943)
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题目链接
这题正解其实是用隔板法,ans(n,k)=C(n+k-1,k-1),因为这题n比较小,所以可以直接用C(n,m)=C(n-1,m-1)+C(n-1,M)打出杨辉三角,时空复杂度都是O(n²)。
然后我用的方法是DP,ans(n,k)=∑(i:0->n)ans(i,k-1), 时间复杂度O(n³)。
附上AC代码:
#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<cstdio>#include<queue>#include<set>#include<vector>#include<map>#include<string>#include<cmath>#define pq priority_queue#define Pi acos(-1.0)#define MAXX 1000000007using namespace std;int gcd(int a,int b){ if(a%b) return gcd(b,a%b); else return b;}int dp[105][105];int main(){ int n,k; for(int i=0;i<=100;i++) dp[i][1]=1; for(int j=2;j<=100;j++) for(int i=0;i<=100;i++) { for(int m=0;m<=i;m++) dp[i][j]+=dp[i-m][j-1]; dp[i][j]%=1000000; } while(scanf("%d%d",&n,&k),n+k!=0) { printf("%d\n",dp[n][k]); } return 0;}
Memory: 0 KB Time: 0 MS
Language: C++ 4.8.2 Result: Accepted
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