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  • codeforces 1428G1 分组背包版本

    include

    include

    include

    include

    include

    include

    include

    using namespace std;
    typedef long long ll;

    const int maxn = 1000000;

    int k,m,num,cntf,cntg;
    int di[maxn];
    ll n,F[10],f[21][maxn],g[maxn],v[1000],w[1000],vg[1000],wg[1000];

    int qsm(int i,int po){
    int res = 1;
    while(po){
    if(po&1) res *= i;
    po >>= 1;
    i *= i;
    }
    return res;
    }

    ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s10+ch-'0'; ch=getchar(); } return sf; }

    int main(){
    num = 0;
    k = read();
    int K = k-1;
    int t = 0;
    while(K - (1<<t) > 0){ // 二进制分组
    di[++num] = 1<<t;
    K -= (1<<t);
    ++t;
    }
    di[++num] = K;
    // for(int i=1;i<=num;++i) printf("%d ",di[i]); printf(" ");

    for(int i=0;i<6;++i) F[i] = read();
    m = read();
    
    
    n = read();
    
    cntg = 0;
    for(int d=1;d<=6;++d){ // 分组 dp 
    	cntf = 0;
    	v[++cntf] = 0, w[cntf] = 0;
    	for(int i=1;i<=3;++i){
    		ll wei = 1ll * (3*i) * qsm(10,d-1);
    		ll val = 1ll * i * F[d-1];
    		
    		for(int j=1;j<=num;++j){
    			v[++cntf] = 1ll * di[j] * val;
    			w[cntf] = 1ll * di[j] * wei;
    		}
    	
    	}
    	
    	memset(f,0,sizeof(f));
    	for(int i=1;i<=cntf;++i){
    		for(int p=1;p<=num;++p){
    			for(int j=n;j>=0;--j){	
    				if(j-w[i]>=0 ) f[p][j] = max(f[p][j],f[p-1][j-w[i]] + v[i]);
    			}
    		}
    	}
    	
    	for(int i=1;i<=n;++i){
    		if(f[num][i] != -1){
    			vg[++cntg] = f[num][i];
    			wg[cntg] = i;
    		}
    	}
    }
    
    vg[++cntg] = 0, wg[cntg] = 0;
    	for(int i=1;i<=cntg;++i){
    			printf("%lld %lld
    ",wg[i],vg[i]);
    		}printf("
    ");	 
    
    memset(g,-1,sizeof(g));
    g[0] = 0;
    for(int i=1;i<=cntg;++i){
    	for(int j=n;j>=0;--j){
    		if(j-wg[i]>=0 && g[j-wg[i]]!=-1) g[j] = max(g[j],g[j-wg[i]] + vg[i]);
    	}
    }
    
    ll ans = 0;
    for(int x=0;x<=n;x++){
    	int tmp = x, o = 1;
    	ll ta = 0;
    	while(tmp > 0){
    		if((tmp%10) % 3 == 0){
    			ta += 1ll * (tmp % 10)/3 * F[o-1];
    		}
    		tmp /= 10;
    		++o;
    	}
    	ans = max(ans,g[n-x]+ ta); //
    }
    
    printf("%lld
    ",ans);
    
    return 0;
    

    }

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  • 原文地址:https://www.cnblogs.com/tuchen/p/13855765.html
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