poj-1285 Combinations, Once Again（DP）

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题目链接：http://poj.org/problem?id=1285

Combinations, Once Again

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1551 Accepted: 522

Description

Given n objects you'd have to tell how many different groups can be chosen if r objects are taken at a time.

Input

Input consists of less than 100 test cases. Each test case begins with two integers n (0 < n <= 50), m (0 <= m <= n). The next line will contain the labels (numbers in the range 1 to n) of the n objects you are to choose from. Two objects with the same label are considered equivalent. Then in the last line for that test case, you'd have m values for r. There will be a single space separating two consecutive numbers in a line. Input is terminated by a test case where n=0, you must not process this test case.

Output

For each test case, print the test case number. And for each query number r, print the number of different groups that can be formed if r objects are taken from the given n objects. You can assume that for all input cases, the output will always fit in a 64-bit unsigned integer and (0<=r<=n).

Sample Input

5 21 2 3 4 52 14 11 2 3 4 20 0

Sample Output

Case 1:105Case 2:6

题意：给一串序列，求一共有多少不同组合。比如样例1，当r=2的时候，有（1,2）（1,3）（1,4）（1,5）（2,3）（2,4）（2,5）（3,4）（3,5）（4,5）一共十种组合。注意当出现重复数字的时候，如果一个组合两个数字都一样，只算一个。比如序列1，1，2 当r=2时有两个解，一个是（1,1）,一个是（1,2）

思路：用dp[i][j]表示从i开始（包括i）r=j时的组合数量。易知dp[i][j]=dp[i+1][j-1]+dp[k][j] k为从i开始第一个不等于i的数（因为dp[k][j]代表不取i，那么[i,k)这个区间内的数都一样，如果取了，就矛盾了。）注意本题不能把数组排序后删去重复的，因为题目可能询问的很大，那么答案是11112223的可能性很大，去掉将很难处理。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[55];
unsigned long long dp[55][55];
int main()
{
int n,m,r;
int q=1;
while(scanf("%d %d",&n,&m)&&(n&&m))
{
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
dp[i][0]=1;
dp[n][1]=1;
int k;
for(int i=n-1; i>=1; i--)
{
for(int j=1; j<=n; j++)
{
dp[i][j]=dp[i+1][j-1];
for(k=i+1;k<=n;k++)
if(a[k]!=a[i])
break;
dp[i][j]+=dp[k][j];
}
}
printf("Case %d:\n",q++);
while(m--)
{
scanf("%d",&r);
printf("%llu\n",dp[1][r]);
}
}
return 0;
}

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