NOIP模拟赛20161114

幸运串

题意：长度为n，字符集大小为m的字符串中有多少不同的不含回文的串

n,m<10^9

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我靠这不就是萌数的DP部分吗

有规律

f[2][j][k]=1

f[i][j][k]=sigma{f[i-1][k][z]|z!=j&&z!=k}

答案就是m*(m-1)*(m-2)^(n-2)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
ll n,m;
inline ll powMod(ll a,ll b){
    ll ans=1;
    for(;b;b>>=1,a=(a*a)%MOD)
        if(b&1) ans=(ans*a)%MOD;
    return ans;
}
int main(){
    freopen("lucky.in","r",stdin); 
    freopen("lucky.out","w",stdout);
    scanf("%lld%lld",&n,&m);
    if(m==1) printf("%d",n==1?1:0); 
    else if(n==1) printf("%d\n",m); 
    else printf("%lld",m*(m-1)%MOD*powMod(m-2,n-2)%MOD);
}

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最优排名

很明显贪心

先v大到小排序

找之前w-v最小的，送她气球，删掉它；送气球之后加入后面又超过自己的

用set超时了，直接用堆就好，只需要删除最大元素

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
const int N=3e5+5;
typedef long long ll;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n;
struct team{
    int id;
    ll w,v;
    bool operator <(const team &r)const{return v>r.v;}
}a[N];

multiset<ll> s;
int main(){
    freopen("rank11.in","r",stdin);
    freopen("rank11.out","w",stdout);
    
    n=read();
    for(int i=1;i<=n;i++) a[i].id=i,a[i].v=read(),a[i].w=read();
    sort(a+1,a+1+n);
    ll p=0,rank;
    for(int i=1;i<=n;i++){
        if(a[i].id==1) break;
        p++;
        s.insert(a[i].w-a[i].v+1);
    }
    rank=p+1;//printf("rank %d\n",rank);
    ll v=a[rank].v,ans=rank,tail=rank+1;
    while(v&&!s.empty()){
        ll now=*s.begin(); //printf("now %d\n",now);
        s.erase(s.begin());
        if(v-now>=0) v-=now,rank--;
        while(v<a[tail].v) {s.insert(a[tail].w-a[tail].v+1);tail++;rank++;}
        ans=min(ans,rank);
    }
    printf("%d",ans);
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int N=3e5+5;
typedef long long ll;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n;
struct team{
    int id;
    ll w,v;
    bool operator <(const team &r)const{return v>r.v;}
}a[N];

priority_queue<ll,vector<ll>,greater<ll> > q;
int main(){
    freopen("rank.in","r",stdin);
    freopen("rank.out","w",stdout);
    
    n=read();
    for(int i=1;i<=n;i++) a[i].id=i,a[i].v=read(),a[i].w=read();
    sort(a+1,a+1+n);
    ll p=0,rank;
    for(int i=1;i<=n;i++){
        if(a[i].id==1) break;
        p++;
        q.push(a[i].w-a[i].v+1);
    }
    rank=p+1;//printf("rank %d\n",rank);
    ll v=a[rank].v,ans=rank,tail=rank+1;
    while(v&&!q.empty()){
        ll now=q.top();q.pop();
        if(v-now>=0) v-=now,rank--;
        while(v<a[tail].v) {q.push(a[tail].w-a[tail].v+1);tail++;rank++;}
        ans=min(ans,rank);
    }
    printf("%d",ans);
}

 

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运输任务

貌似是树链剖分，不会

能骗20到50分吧