SDOI 2016 Round1 Day2

生成魔咒

/*
后缀数组+双向链表
参照：https://blog.csdn.net/clove_unique/article/details/53911757
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+5;
int n,cnt,c[N],ans[N],s[N],t[N],rank[N],trank[N],sa[N],tsa[N],h[N],nxt[N],pre[N];
long long res;
void DA(){
    int p,maxx=cnt;
    memset(c,0,sizeof c);
    for(int i=1;i<=n;i++) c[rank[i]=s[i]]++;
    for(int i=1;i<=maxx;i++) c[i]+=c[i-1];
    for(int i=n;i;i--) sa[c[rank[i]]--]=i;
    trank[sa[1]]=p=1;
    for(int i=2;i<=n;i++){
        if(rank[sa[i]]!=rank[sa[i-1]]) p++;
        trank[sa[i]]=p;
    }
    for(int i=1;i<=n;i++) rank[i]=trank[i];
    for(int k=1;p<n;k<<=1,maxx=p){
        p=0;
        for(int i=n-k+1;i<=n;i++) tsa[++p]=i;
        for(int i=1;i<=n;i++) if(sa[i]>k) tsa[++p]=sa[i]-k;
        memset(c,0,sizeof c);
        for(int i=1;i<=n;i++) trank[i]=rank[tsa[i]];
        for(int i=1;i<=n;i++) c[trank[i]]++;
        for(int i=1;i<=maxx;i++) c[i]+=c[i-1];
        for(int i=n;i;i--) sa[c[trank[i]]--]=tsa[i];
        trank[sa[1]]=p=1;
        for(int i=2;i<=n;i++){
            if(rank[sa[i]]!=rank[sa[i-1]]||rank[sa[i]+k]!=rank[sa[i-1]+k]) p++;
            trank[sa[i]]=p;
        }
        for(int i=1;i<=n;i++) rank[i]=trank[i];
    } 
    for(int i=1,k=0;i<=n;i++){
        int j=sa[rank[i]-1];
        while(s[i+k]==s[j+k]) k++;
        h[rank[i]]=k;if(k>0) k--;
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&s[i]),t[i]=s[i];
    for(int i=1;i<=n/2;i++) swap(s[i],s[n-i+1]);
    sort(t+1,t+n+1);
    cnt=unique(t+1,t+n+1)-(t+1);
    for(int i=1;i<=n;i++) s[i]=lower_bound(t+1,t+cnt+1,s[i])-t;
    DA();
    for(int i=1;i<=n;i++){
        nxt[i]=i+1;
        pre[i]=i-1;
    }
    for(int i=1;i<=n;i++){
        int k=rank[i];
        ans[i]=n-i+1-max(h[k],h[nxt[k]]);
        h[nxt[k]]=min(h[k],h[nxt[k]]);
        nxt[pre[k]]=nxt[k];
        pre[nxt[k]]=pre[k];
    }
    for(int i=n;i;i--){
        res+=ans[i];
        printf("%lld
",res);
    }
    return 0;
}

排列计数

/*
ans=C(n,m)*dp(n-m);|dp(i)表示i的错排数
*/
#include<cstdio>
using namespace std;
typedef long long ll;
inline ll read(){
    ll x=0;char ch=getchar();
    while(ch<'0'||ch>'9'){ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
 
#define FRE(name) freopen(#name".in","r",stdin);freopen(#name".out","w",stdout);
#define fre(name) freopen(#name".txt","r",stdin);
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
 
const ll S=1e6;
const ll N=S+5;
const ll mod=1e9+7;
ll cas,n,m,ans,dp[N],fz[N],fm[N];
ll fpow(ll a,ll p){
    ll res=1;
    for(;p;p>>=1,a=a*a%mod) if(p&1) res=res*a%mod;
    return res;
}
void first(){
    dp[0]=1;dp[1]=0;
    fz[1]=fz[0]=fm[1]=fm[0]=1;
    for(ll i=2;i<=S;i++) dp[i]=(i*dp[i-1]%mod+((i&1)?-1:1)+mod)%mod;
    for(ll i=2;i<=S;i++) fz[i]=fz[i-1]*i%mod;
    fm[S]=fpow(fz[S],mod-2);
    for(ll i=S-1;i>=1;i--) fm[i]=fm[i+1]*(i+1)%mod;
}
void solve(){
    for(cas=read();cas--;){
        n=read();m=read();
        ans=(((fz[n]*fm[m])%mod*fm[n-m])%mod*dp[n-m])%mod;
        printf(LL "
",ans);
    }
}
int main(){
    //FRE(permutation);
    first();
    solve();
    return 0;
}

征途

/*
ans=m(x_1^2+X_2^2+……+x_m^2)-(x_1+X_2+……+x_m)^2
令dp[i][j]表示前i个数已经选择到了x_j的最小值；这样就变成了一个区间dp
我懒，就不写单调队列优化了
*/
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
 
#define pf(x) ((x)*(x))
#define fro __attribute__((optimize("O2")))
#define FRE(name) freopen(#name".in","r",stdin);freopen(#name".out","w",stdout);
#define fre(name) freopen(#name".txt","r",stdin);
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
 
const int N=3005;
int n,m,a[N],p[N][N];
ll ans,sum[N],dp[N][N];
 
fro int main(){
    //FRE(journey);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),sum[i]=sum[i-1]+a[i];
    memset(dp,0x3f3f3f3f,sizeof dp);
    dp[0][0]=0;
    for(int j=1;j<=m;j++){
        for(int i=j;i<=n;i++){
            for(int k=p[i-1][j];k<i;k++){
                if(dp[i][j]>dp[k][j-1]+pf(sum[i]-sum[k])){
                    dp[i][j]=dp[k][j-1]+pf(sum[i]-sum[k]);
                    p[i][j]=k;
                }
            }
        }
    }
    ans=dp[n][m]*(ll)m-pf(sum[n]);
    printf(LL,ans);
    return 0;
}