POJ-3046 Ant Counting
题面
Problem 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.
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.
题意
找含k元set的个数。
题解来自：http://blog.csdn.net/sdj222555/article/details/10440021
很容易想到dp[i][j]表示选到第i种，已经放了几个元素。但是很不争气的写挂了。
书上指的优化递推关系式 大概就是指的是用前缀和。。
可以学习到的地方是上面链接的题解用了两个指针指向数组，然后就不会搞错，值得借鉴233.
真觉得poj该升级辣QWQ。。。
代码
#include <iostream>
#include <algorithm>
using namespace std;
int dp[2][100010], num[1111];
int sum[100010], up[1111];
int t,n,l,r,x;
const int MOD=1000000;
int ans;
int main()
{
cin>>t>>n>>l>>r;
for (int i=1;i<=n;i++)
{
cin>>x;
num[x]++;
}
for (int i=1;i<=t;i++)
up[i]=up[i-1]+num[i];
dp[0][0]=1;
int *pre = dp[0], *nxt = dp[1];
for (int i=1;i<=t;i++)
{
sum[0]=pre[0];
for (int j=1;j<=up[i];j++)
sum[j]=(sum[j-1]+pre[j]) % MOD;
for (int j=0;j<=up[i];j++)
{
int tmp=max(0,j-num[i]);
nxt[j]=tmp==0?sum[j]:(sum[j]-sum[tmp-1])%MOD;
nxt[j]%=MOD;
}
swap(nxt,pre);
}
for(int i=l;i<=r;i++)
ans=(ans+pre[i])%MOD;
cout<<ans<<endl;
return 0;
}