传送门
按照题目的意思,我们可以把每一行看成一组,然后把每一行处理出来当做分组背包来做,但如果直接暴力去做的话值域太大,所以我们可以按照这题的特性,因为k很小,所以我们按照余数来区分这些值域,处理好每一行的复杂度是(35×70×70×35),之后的分组背包dp复杂度是(70×70×70),足以过这题.
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 70*35;
const int INF = 0x8f8f8f8f;
inline void read(int &a){
int x = 0,f=1;char ch = getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){
x=x*10+ch-'0';ch=getchar();}
a = x*f;
}
int a[90],dp[4910],bag[75][75],f[4910],dpp[75][75];
int num,nn,m,kk;
int maxsum(int row){
mem(dp,0);
mem(f,0x3f);
dp[0] = 1;
f[0] = 0;
fir(i,1,m){
afir(j,M,a[i]){
if(!dp[j] && dp[j-a[i]] && f[j-a[i]] < num){
dp[j] = 1;
f[j] = f[j-a[i]] + 1;
}
else if(dp[j])
f[j] = min(f[j],f[j-a[i]]+1);
}
}
fir(i,1,4900)
if(dp[i])
bag[row][i%kk] = i;
return 0;
}
int main(){
int n;
cin >> n >> m >> kk;
num = m/2;
int ans = 0;
mem(bag,0x8f);
fir(i,1,n){
fir(j,1,m)
cin >> a[j];
maxsum(i);
}
mem(dpp,0x8f);
dpp[0][0] = 0;
fir(i,1,n){
fir(j,0,kk-1){
dpp[i][j] = dpp[i-1][j];
fir(k,0,kk-1){
if(bag[i][(k-j+kk)%kk] == INF) continue;
dpp[i][j] = max(dpp[i][j],dpp[i-1][k]+bag[i][(k-j+kk)%kk]);
}
}
}
fir(i,1,n) ans = max(ans,dpp[i][0]);
cout << ans;
return 0;
}