BZOJ 1197 花仙子的魔法(递推)
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数学归纳法。
dp[i][j]=dp[i][j-1]+dp[i-1][j-1].
![](https://image.cha138.com/20210609/46bc77072a4848afa4364863aaab2b73.jpg)
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 100000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==\'-\') flag=1; else if(ch>=\'0\'&&ch<=\'9\') res=ch-\'0\'; while((ch=getchar())>=\'0\'&&ch<=\'9\') res=res*10+(ch-\'0\'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(\'-\'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+\'0\'); } const int N=10005; //Code begin... LL dp[20][105]; void init() { FOR(i,1,100) dp[1][i]=2*i; FOR(i,1,15) dp[i][1]=2; FOR(i,2,15) FOR(j,2,100) dp[i][j]=dp[i][j-1]+dp[i-1][j-1]; } int main () { init(); int n, m; scanf("%d%d",&m,&n); printf("%lld\\n",dp[n][m]); return 0; }
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