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mb62d7cd05038ca 2017-09-01 18:14:02 博主文章分类：比赛题解 ©著作权

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BZOJ 4971 记忆中的背包

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
ll C[40][40];
ll g[MAXN][40];
ll p[MAXN][40];
ll f[MAXN][40]; //w ,c(I,>)
void prework() {
    int n=30;
    C[0][0]=1;
    For(i,n) {
        C[i][0]=1;
        For(j,i) {
            C[i][j]=(C[i-1][j-1]+C[i-1][j]);
        } 
    }
    MEMI(f)
    ForD(i,30) {
        f[0][i]=0;
        For(j,20000) Rep(x,(i+1)/2+1) if (j>=C[i][x]){
            if (f[j][i]> f[j-C[i][x]][i] + 1 ) {
                f[j][i]=f[j-C[i][x]][i] + 1;
                g[j][i]=j-C[i][x];  
                p[j][i]=x;
            }
        }
    }
}
ll w,mo,k;
int ans[MAXN];
void work() {
    if (!k) {
        puts("1");
        cout<<w-1<<endl;
    }else if (k<=40) {
        cout<<k<<endl;
        For(i,k-1) cout<<w<<' ';cout<<w<<endl;
    }else {

        For(i,30) {
            if (f[k][i]+i<=40) {
                printf("%d\n",f[k][i]+i);
                int cnt=0;
                while(k) {
                    ans[++cnt]=w-p[k][i] ;
                    k=g[k][i];
                }
                while(i--) ans[++cnt]=1;
                PRi(ans,cnt);
                return;
            }
        }
    }
}
int main()
{
//  freopen("bzoj4971.in","r",stdin);
//  freopen(".out","w",stdout);
    int T=read();
    prework();
    while(T--) {
        scanf("%lld%lld%lld",&w,&mo,&k);
        work();
    }


    return 0;
}

BZOJ 4972 小Q的方格纸

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef unsigned int uint;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
int n,m,q; 
unsigned int A,B,C;
uint a[MAXN][MAXN];
inline unsigned int rng61(){
    A ^= A << 16;
    A ^= A >> 5;
    A ^= A << 1;
    unsigned int t = A;
    A = B;
    B = C;
    C ^= t ^ A;
    return C;
}
uint s[MAXN][MAXN];
uint t[MAXN][MAXN];
void prework() {
    MEM(s) MEM(t)
    For(i,n) For(j,m) s[i][j]=s[i][j-1]+s[i-1][j]-s[i-1][j-1]+a[i][j];
    For(i,n) For(j,m) {
        t[i][j]=t[i-1][j]+a[i][j];
    }
    For(i,n) ForD(j,m) {
        t[i][j]+=t[i-1][j+1];
    }
} 
uint calc(int x,int y,int k) {
    uint p=s[x][y] - s[x][y-k] ;
    uint l=t[x-1][y-k+1]-t[x-k-1][y+1];
    return p-l;
}
uint f[3000000+10];
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);
    int i,j;
    uint x,y,k;
    scanf("%d%d%d%u%u%u", &n, &m, &q, &A, &B, &C);
    for(i = 1; i <= n; i++)
        for(j = 1; j <= m; j++)
            a[i][j] = rng61();
//    PRi2D(a,n,m)
    prework();
//    cout<<a[3][2]+a[3][1]+a[2][2];
    for(i = 1; i <= q; i++){
        x = rng61() % n + 1;
        y = rng61() % m + 1;
        k = rng61() % min(x, y) + 1;
//        cout<<x<<' '<<y<<' '<<k<<endl;
        f[i]=calc(x,y,k);
//      cout<<f[i]<<endl;
    }
    uint p=1,ans=0;
    ForD(i,q) {
        ans+=p*f[i];
        p*=(uint)233;
    }
    printf("%u\n",ans);


    return 0;
}

BZOJ 4974 字符串大师

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
bool b[26];
char s[MAXN];
int f[MAXN];
int main()
{
//  freopen("D.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read();
    For(i,n) f[i]=read();
    f[0]=0; int p=0;
    For(i,n) {
        if (i==1) s[i]='a';
        else if (f[i]!=i) {
            s[i]=s[i- f[i] ];
        }
        else {
            int p=i-1;
            p-=f[p];
            MEM(b) 
            while(p) {
                b[s[p+1]-'a']=1;
                p-=f[p];
            }
            b[s[p+1]-'a']=1;
            Rep(j,26) {
                if (!b[j]) {
                    s[i]=j+'a';
                    break;
                }
            }
        }
    }
    printf("%s",s+1);
    return 0;
}

BZOJ 4975 区间翻转

#include<bits/stdc++.h> 
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int a[1234];
int main()
{
//  freopen("bzoj4975.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read();
    For(i,n) {
        a[i]=read();    
    }
    int p=0;
    For(i,n) Fork(j,i+1,n) if (a[i]<a[j]) p^=1;
    puts(!p?"T":"Q");

    return 0;
}

BZOJ 4976 宝石镶嵌

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#define
#define

typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int f[1<<17];
int a[123456];
int main()
{
//  freopen("bzoj4976.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read(),k=read();
    For(i,n) a[i]=read();
    int mx=*max_element(a+1,a+1+n);
    if (n>=k+17) {
        int ans=0;
        For(i,n) ans|=a[i];
        cout<<ans<<endl;
    } else {
        MEMI(f) f[0]=0;
        int S=1<<17;
        For(i,n) {
            Rep(st,S) gmin(f[st| a[i]] , f[st]+1); 
        }
        int ans=0;
        Rep(st,S) if (f[st]<=n-k) ans=st;
        cout<<ans<<endl;
    }
    return 0;
}