2014
03-01

# Parliament Seat

It is election time. V voters attend the election, each casting their vote for one of N political parties. Mofficials will be elected into the parliament.
The conversion from votes to parliament seats is done using the D’Hondt method with a 5% threshold.More precisely, suppose that the parties are numbered 1 through N and that they receive V1, V2, …, VN votes. Parliament seats are allocated as follows:
1. All parties that receive strictly less than 5% of V votes are erased from the list of parties.
2. The parliament is initially empty i.e. every party has zero seats allocated.
3. For each party P, the quotient QP=VP/(SP+1) is calculated, where VP is the total number of votes received by party P, and SP is the number of seats already allocated to party P.
4. The party with the largest quotient QP is allocated one seat. If multiple parties have the same largest quotient, the lower numbered party wins the seat.
5. Repeat steps 3 and 4 until the parliament is full.
The votes are being counted and only part of the V votes has been tallied. It is known how many votes each party has received so far.Write a program that calculates for each party, among all possible outcomes of the election after all V votes are counted, the largest and smallest number of seats the party wins.

The first line contains the integers V, N and M (1 ≤ V ≤ 10,000,000, 1 ≤ N ≤ 100, 1 ≤ M ≤ 200), the numbers of votes, parties and seats in the parliament.
The second line contains N integers � how many votes (of those that have been counted) each party got. The sum of these numbers will be at most V.

The first line contains the integers V, N and M (1 ≤ V ≤ 10,000,000, 1 ≤ N ≤ 100, 1 ≤ M ≤ 200), the numbers of votes, parties and seats in the parliament.
The second line contains N integers � how many votes (of those that have been counted) each party got. The sum of these numbers will be at most V.

20 4 5
4 3 6 1
100 3 5
30 20 10

3 3 3 2
1 0 1 0
4 3 3
1 1 0

Hint
In the first example 14 votes have been tallied and 6 are yet to be counted. To illustrate one possible outcome, suppose that the first party receives 2 of those 6 votes,
the second none, the third 1 vote and the fourth 3 votes. The parties' totals are 6, 3, 7 and 4 votes. All parties exceeded the 5% threshold.
Seats are allocated as follows:
1. The quotients are initially 6/1, 3/1, 7/1 and 4/1; the largest is 7/1 so party 3 wins a seat.
2. The quotients are 6/1, 3/1, 7/2 and 4/1; the largest is 6/1 so party 1 wins a seat.
3. The quotients are 6/2, 3/1, 7/2 and 4/1; the largest is 4/1 so party 4 wins a seat.
4. The quotients are 6/2, 3/1, 7/2 and 4/2; the largest is 7/2 so party 3 wins a seat.
5. The quotients are 6/2, 3/1, 7/3 and 4/2; parties 1 and 2 are tied with quotients 6/2 and 3/1,but party 1 is lower numbered so it wins the last seat.

In this outcome, the numbers of seats won by the parties are 2, 0, 2 and 1. Since it is possible for the second party not to win any seats, the second number
on the second line of output is zero.


Fi[j] = Fi-1[j] (Vi>j)
Fi[j] = Fi-1[j] + Fi-1[j-Vi](j>=Vi)

*C)。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define Maxn 4000
using namespace std;
int v[Maxn],dp[10100],ans,Min,Sum;
int main()
{
int t,n,m,i,j,Case=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
Sum=0;
Min=0x7fffffff;
for(i=1;i<=n;i++)
{
scanf("%d",v+i);
Min=min(Min,v[i]);
Sum+=v[i];
}
ans=0;
int sum=0,f=0,k;
sort(v+1,v+1+n);
for(i=1;i<=n;i++)
{
memset(dp,0,sizeof(dp));
dp[sum]=1;
for(j=i+1;j<=n;j++)
for(k=m;k>=sum+v[j];k--)
dp[k]=dp[k]+dp[k-v[j]];
for(j=m;j>=m-v[i]+1;j--)
if(j>=sum) ans+=dp[j];
sum+=v[i];
}
if(Sum<=m)
printf("%d 1\n",++Case);
else
if(Min>m)
printf("%d 0\n",++Case);
else
printf("%d %d\n",++Case,ans);
}
return 0;
}

1. L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-1]）这个地方也也有笔误
应改为L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-2]）

2. 有限自动机在ACM中是必须掌握的算法，实际上在面试当中几乎不可能让你单独的去实现这个算法，如果有题目要用到有限自动机来降低时间复杂度，那么这种面试题应该属于很难的级别了。