CodeForces-1348E Phoenix and Berries

传送门：CodeForces-1348E

思路分析

设dp[i][j]表示到第i棵树，剩了j个红色的最大筐数，蓝色的数量可以通过红色计算出来
枚举dp[i-1][j]来更新状态，对于当前树有两种情况：

1. 选择放进同树筐内，枚举s表示放s个红色
2. 不放同树筐

具体转移见代码部分

这题比较毒瘤卡了ll取模

样例输入

2 5
2 1 
1 3

样例输出

0

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 555;

ll dp[N][N],a[N],b[N]; 

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,k;
    read(n,k);
    for(int i=1;i<=n;i++) read(a[i],b[i]);
    mem(dp,-1);
    dp[0][0]=0;
    ll sum=0;
    for(int i=1;i<=n;i++){
        sum+=a[i]+b[i];
        for(int j=0;j<k;j++){
            if(dp[i-1][j]<0) continue;
            for(int s=1;s<k&&s<=a[i];s++){
                if(k-s>b[i]) continue;
                int sum1=j+a[i]-s;
                int sum2=sum-dp[i-1][j]*k-k-sum1;
                dp[i][sum1%k]=max(dp[i][sum1%k],dp[i-1][j]+sum1/k+sum2/k+1);
            }
            int sum1=j+a[i];
            int sum2=sum-dp[i-1][j]*k-sum1;
            dp[i][sum1%k]=max(dp[i][sum1%k],dp[i-1][j]+sum1/k+sum2/k);
        }
    }
    
    ll ans=0;
    for(int j=0;j<k;j++) ans=max(dp[n][j],ans);
    cout<<ans<<endl;




    return 0;
}