ZOJ - 2402 DP方案数

题意:给出m,序列第i位是第i-1位的至少2倍大,的求长度为n且每一位范围均在1-m的序列方案数

对求方案数做不到信手拈来的感觉,需要加强

用简单的预处理和最优子结构能优化到很不错的效率了

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e3+11;

const int oo = 0x3f3f3f3f;

const double eps = 1e-7;

typedef long long ll;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

ll dp[12][6666];//长度为i的第i位大小为j的方案数 //ans:dp[n][1..m]

//dp[i][j]=Σdp[i-1][1...j/2]

ll sum[6666];

int main(){

	int T=read(),kase=0;

	rep(i,1,2000) dp[1][i]=1;

	rep(i,2,10){

		rep(j,1,2000){

			sum[j]=sum[j-1]+dp[i-1][j];

			dp[i][j]=sum[j/2];

		}

	}

	while(T--){

		int n=read();

		int m=read();

		ll ans=0;

		rep(i,1,m) ans+=1ll*dp[n][i];

		printf("Case %d: n = %d, m = %d, # lists = %lld\n",++kase,n,m,ans);

	}

	return 0;

}