【BZOJ】1630: [Usaco2007 Demo]Ant Counting（裸dp/dp/生成函数）

http://www.lydsy.com/JudgeOnline/problem.php?id=1630

题意，给你n种数，数量为m个，求所有的数组成的集合选长度l～r的个数

后两者待会写。。

裸dp其实应该会tle的额，但是数据弱？

d[i][j]表示前i种j长度的数量

d[i][j]=sum{d[i-1][j-k]} 1<=k<=a[i]

会爆mle。但是发现这是裸动态数组。。

注意顺序即可

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }



const int N=1005, md=1e6;

int a[N], n, m, f[N*100], l, r, ans;



int main() {

	read(n); read(m); read(l); read(r);

	for1(i, 1, m) ++a[getint()];

	for1(i, 0, a[1]) f[i]=1;

	for1(i, 2, n) {

		for3(j, r, 0)

			for1(k, 1, a[i]) if(j<k) break; else f[j]=(f[j]+f[j-k])%md;

	}

	for1(i, l, r) ans=(ans+f[i])%md;

	printf("%d", ans);

	return 0;

}

 

然后是前缀和一优化

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }



const int N=1005, md=1e6;

int a[N], n, m, f[N*100], sum[N*100], l, r, ans;



int main() {

	read(n); read(m); read(l); read(r);

	for1(i, 1, m) ++a[getint()];

	f[0]=1;

	for1(i, 1, n) {

		sum[0]=1;

		for1(j, 1, r) sum[j]=(sum[j-1]+f[j])%md;

		for3(j, r, 1)

			if(j<=a[i]) f[j]=sum[j]%md;

			else f[j]=(sum[j]-sum[j-a[i]-1])%md;

	}

	for1(i, l, r) ans=(ans+f[i])%md;

	printf("%d", ans);

	return 0;

}

 

 

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Description

Bessie was poking around the ant hill one day watching the ants march to and fro while gathering food. She realized that many of the ants were siblings, indistinguishable from one another. She also realized the sometimes only one ant would go for food, sometimes a few, and sometimes all of them. This made for a large number of different sets of ants! Being a bit mathematical, Bessie started wondering. Bessie noted that the hive has T (1 <= T <= 1,000) families of ants which she labeled 1..T (A ants altogether). Each family had some number Ni (1 <= Ni <= 100) of ants. How many groups of sizes S, S+1, ..., B (1 <= S <= B <= A) can be formed? While observing one group, the set of three ant families was seen as {1, 1, 2, 2, 3}, though rarely in that order. The possible sets of marching ants were: 3 sets with 1 ant: {1} {2} {3} 5 sets with 2 ants: {1,1} {1,2} {1,3} {2,2} {2,3} 5 sets with 3 ants: {1,1,2} {1,1,3} {1,2,2} {1,2,3} {2,2,3} 3 sets with 4 ants: {1,2,2,3} {1,1,2,2} {1,1,2,3} 1 set with 5 ants: {1,1,2,2,3} Your job is to count the number of possible sets of ants given the data above. //有三个家庭的ANT，共五只，分别编号为1,2,2,1,3，现在将其分为2个集合及3集合，有多少种分法

Input

* Line 1: 4 space-separated integers: T, A, S, and B * Lines 2..A+1: Each line contains a single integer that is an ant type present in the hive

Output

* Line 1: The number of sets of size S..B (inclusive) that can be created. A set like {1,2} is the same as the set {2,1} and should not be double-counted. Print only the LAST SIX DIGITS of this number, with no leading zeroes or spaces.

Sample Input

3 5 2 3
1
2
2
1
3

INPUT DETAILS:

Three types of ants (1..3); 5 ants altogether. How many sets of size 2 or
size 3 can be made?

Sample Output

10

OUTPUT DETAILS:

5 sets of ants with two members; 5 more sets of ants with three members

HINT

Source

Silver