Codeforces#302-C. Writing Code-DP/滚动数组
http://codeforces.com/contest/544/problem/C
题意：
给你n,m,k,mod，表示有n个程序员，要求写m行代码，代码的bugs不超过k个，求方案数模mod
接下来是一个n个数的数组，表示第i个程序员写一行代码会有tm[i]个 bugs 【被黑得好凄凉的感觉....】
n,m,k都是 【1，500】;
如果你选了X个程序员，你只需要保证一共写了m行代码，可以有y个人完全不写....
思路：
dp解决...主要是就转移方程有点麻烦..
dp[ i ][ j ][ k ]表示前i个程序员写了j行代码且bug不超过k
对其dp[ i ][ j ][ k ]分两种情况讨论，
1】 第i个程序员一行代码都没写
2】第i个程序员至少写了1行
情况1】，如果第i个程序员一行都不写，显然就是前i-1个人写完了所有代码，也就是dp[ i ][ j ][ k ]=dp[ i-1 ][ j ][ k ]种
情况2】，如果第i个程序员至少写了1行，我们怎么表示呢
难道是这样？？ int sum=0;
for (line =1;line<= j;line ++)
{
sum+=dp[ i-1 ] [ j-line ] [ k-line*a[ i ] ];
}
显然如果这样，dp就是O(n^4),显然不行....
我们再看看 以前计算得到的结果是否能被利用：
我们可以发现dp[ i ] [ j-1 ] [ k-a[i] ] 是在dp[ i ][ j ][ k ]之前已经计算过了的，
其意义是前i个程序员里面 一共写了j-1行代码，得到的bugs是k-a[i]，隐藏意思就是(这个方案里,第i个程序员可能写过的代码行数为 【0，j-1】)
显然，对于dp[ i ] [ j-1 ] [ k-a[i] ] 下的每一种方案得到的代码，只要再加上一行第i 个程序员写的代码，那么行数就会变为J，bugs数就变为k,而第i个程序员写过的代码
行数就从【0，j-1】变为【1，j】，也就是 情况2】 【第i个程序员至少写了1行】,所以dp[ i ] [ j-1 ] [ k-a[i] ] 就是该情况下的方案数
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从而我们得到了转移方程 dp[ i ] [ j-1 ] [ k-a[i] ]=dp[ i-1 ][ j ][ k ] + dp[ i ] [ j-1 ] [ k-a[i] ] ;
显然开三位数组爆内存啦。。注意到状态方程的 【i】 只用到了i,i-1，那么我们可以用滚动数组，或者直接不用也行了。。
上述方程就转化为了
dp[j][k]=(dp[j][k]+dp[j-1][k-tm[i]]); // 第一个dp[j][k]是指dp[i][j][k]，第二个dp[j][k]由于尚未更新,其实是上一次的结果 也就是dp[i-1][j][k],第三个j-1已经更新过了,所以就是dp[i]的j-1
//未优化空间方程 for (i=1;i<=n;i++) for (j=1;j<=m;j++) for (k=0;k<=b;k++) if (k>=tm[i]) dp[i][j][k]= (dp[i-1][j][k]+dp[i][j-1][k-tm[i]])%mod; else dp[i][j][k]=dp[i-1][j][k];
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;
int dp[505][505];
int tm[505];
int main()
{
int n,m,b,mod;
int k,i,j;
scanf("%d%d%d%d",&n,&m,&b,&mod);
for (i=1;i<=n;i++)
{
scanf("%d",&tm[i]);
}
//sort(tm+1,tm+1+n);
dp[0][0]=1;
for (i=1;i<=n;i++)
{
for (j=1;j<=m;j++)
{
for (k=tm[i];k<=b;k++)
{
dp[j][k]=(dp[j][k]+dp[j-1][k-tm[i]])%mod;
}
}
}
int sum=0;
for (i=0;i<=b;i++)
sum=(sum+dp[m][i])%mod;
printf("%d\n" ,sum);
return 0;
}
//未优化空间方程 <pre class="source prettyprint" style="margin-top: 0px; margin-bottom: 0px; padding: 2px; font-size: 13.2px; border: 1px solid rgb(136, 136, 136); background-color: rgb(255, 255, 255);"><span class="kwd" style="color: rgb(0, 0, 136);">for</span><span class="pln"> </span><span class="pun" style="color: rgb(102, 102, 0);">(</span><span class="pln">i</span><span class="pun" style="color: rgb(102, 102, 0);">=</span><span class="lit" style="color: rgb(0, 102, 102);">1</span><span class="pun" style="color: rgb(102, 102, 0);">;</span><span class="pln">i</span><span class="pun" style="color: rgb(102, 102, 0);"><=</span><span class="pln">n</span><span class="pun" style="color: rgb(102, 102, 0);">;</span><span class="pln">i</span><span class="pun" style="color: rgb(102, 102, 0);">++)</span><span class="pln"> </span><span class="kwd" style="color: rgb(0, 0, 136);">for</span><span class="pln"> </span><span class="pun" style="color: rgb(102, 102, 0);">(</span><span 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