Contest1828 - 2019年我能变强组队训练赛热身赛

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1006;
const ll mod=1000000007;
int T;
int n,m;
ll k;
ll dp[1005][6],a[maxn];
ll gdp[1005][6];
//ll ans;
char f[maxn];
int main() {
    //freopen("1.txt","r",stdin);
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%lld",&n,&m,&k);
        for(int i=1;i<=n;++i){
            scanf("%lld",a+i);
        }
        scanf("%s",f+1);
        for(int i=0;i<=n;++i){
            for(int j=0;j<=m;++j){
                dp[i][j]=-1000000000000009;
                gdp[i][j]=1000000000000009;
            }
        }
        dp[0][0]=k;
        gdp[0][0]=k;
        for(int i=1;i<=n;++i){
            for(int j=max(0,m-n+i);j<=min(i,m);++j){
                dp[i][j]=dp[i-1][j];
                gdp[i][j]=gdp[i-1][j];
                if(j==0)continue;
                if(f[j]==‘+‘){
                    dp[i][j]=max(dp[i][j],max(dp[i-1][j-1]+a[i],gdp[i-1][j-1]+a[i]));
                    gdp[i][j]=min(gdp[i][j],min(dp[i-1][j-1]+a[i],gdp[i-1][j-1]+a[i]));
                }
                else if(f[j]==‘-‘){
                    dp[i][j]=max(dp[i][j],max(dp[i-1][j-1]-a[i],gdp[i-1][j-1]-a[i]));
                    gdp[i][j]=min(gdp[i][j],min(dp[i-1][j-1]-a[i],gdp[i-1][j-1]-a[i]));
                }
                else if(f[j]==‘*‘){
                    dp[i][j]=max(dp[i][j],max(dp[i-1][j-1]*a[i],gdp[i-1][j-1]*a[i]));
                    gdp[i][j]=min(gdp[i][j],min(dp[i-1][j-1]*a[i],gdp[i-1][j-1]*a[i]));
                }
                else{
                    dp[i][j]=max(dp[i][j],max(dp[i-1][j-1]/a[i],gdp[i-1][j-1]/a[i]));
                    gdp[i][j]=min(gdp[i][j],min(dp[i-1][j-1]/a[i],gdp[i-1][j-1]/a[i]));
                }
                //ans=max(ans,dp[i][j]);
                //cout<<i<<‘ ‘<<j<<"dpij"<<‘ ‘<<dp[i][j]<<endl;
            }
        }
        printf("%lld\n",dp[n][m]);
    }
    return 0;
}


#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
ll n,m;

struct node
{
    ll a[9][9];
    node()
    {
        memset(a,0,sizeof(a));
    }
    node operator*(const node &b) const
    {
        node res;
        for (int i = 0; i < 9; i++)
        {
            for (int j = 0; j < 9; j++)
            {
                res.a[i][j] = 0;
                for (int k = 0; k < 9; k++)
                {
                    res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]%mod) % mod;
                }
            }
        }
        return res;
    }
};

node pow(node b,ll c)
{
    node res;
    for (int i=0; i<9; i++)
    {
        res.a[i][i]=1;
    }
    while (c)
    {
        if (c & 1)
        {
            res = res * b;
        }
        c >>= 1;
        b = b * b;
    }
    return res;
}
int a[9][9]={
    0,1,1,0,0,0,0,0,0,
    0,0,0,1,1,1,0,0,0,
    0,0,0,0,0,0,1,0,1,
    1,1,1,0,0,0,0,0,0,
    0,0,0,1,0,1,0,0,0,
    0,0,0,0,0,0,0,1,1,
    1,1,0,0,0,0,0,0,0,
    0,0,0,1,1,0,0,0,0,
    0,0,0,0,0,0,1,1,0};
node f;
int main()
{
    for (int i=0; i<9; i++)
    {
        for (int j=0; j<9; j++)
            f.a[i][j]=a[i][j];
    }
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%lld",&n);
        if (n==1){
            printf("3\n");
            continue;
        }
        if (n==2){
            printf("9\n");
            continue;
        }
        node ans=pow(f,n-2);
        ll ans1=0;
        for (int i=0; i<9; i++)
        {
            for (int j=0; j<9; j++)
            {
                ans1=(ans1+ans.a[i][j])%mod;
            }
        }
        printf("%lld\n",ans1);
    }
}
#include<bits/stdc++.h>

using namespace std;

#define rep(i,a,n) for(int i=a;i<n;i++)

#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;

const ll mod = 1e9+7;

ll powmod(ll a,ll b) {
    ll ret = 1;
    a%=mod;
    while(b) {
        if(b&1) ret = ret*a%mod;
        a = a*a%mod;
        b>>=1;
    }
    return ret;
}
int _,n;

namespace linear_seq {
const int N = 10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;

void mul(ll *a,ll *b,int k) {
    for(int i=0; i<k+k; i++)
        _c[i] = 0;
    for(int i=0; i<k; ++i)
        if(a[i])
            for(int j=0; j<k; j++)
                _c[i+j] = (_c[i+j]+a[i]*b[j])%mod;

    for(int i=k+k-1; i>=k; i--)
        if(_c[i])
            for(int j=0; j<(int) Md.size(); j++)
                _c[i-k+Md[j]] = (_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
    for(int i=0; i<k; i++)
        a[i] = _c[i];
}

int solve(ll n,VI a,VI b) {
    ll ans = 0,pnt = 0;
    int k = SZ(a);
    for(int i=0; i<k; i++)
        _md[k-1-i] = -a[i];
    _md[k] = 1;
    Md.clear();
    for(int i=0; i<k; i++)
        if(_md[i]!=0)
            Md.push_back(i);
    for(int i=0; i<k; i++)
        res[i] = base[i] = 0;
    res[0] = 1;
    while((1ll<<pnt)<=n)
        pnt++;

    for(int p = pnt; p>=0; p--) {
        mul(res,res,k);
        if((n>>p)&1) {
            for(int i=k-1; i>=0; i--) res[i+1] = res[i];
            res[0]  = 0;
            for(int j=0; j<(int) Md.size(); ++j)
                res[Md[j]] = (res[Md[j]]-res[k]*_md[Md[j]])%mod;
        }
    }
    rep(i,0,k) ans = (ans+res[i]*b[i])%mod;
    if(ans<0) ans+=mod;
    return ans;
}

VI BM(VI s) {
    VI C(1,1),B(1,1);
    int L = 0,m = 1,b = 1;
    for(int n=0; n<(int)s.size(); ++n) {
        ll d = 0;
        for(int i=0; i<L+1; i++)
            d = (d+(ll)C[i]*s[n-i])%mod;
        if(d==0) ++m;
        else if(2*L<=n) {
            VI T=C;
            ll c = mod-d*powmod(b,mod-2)%mod;
            while(SZ(C)<SZ(B)+m)
                C.push_back(0);
            for(int i=0; i<(int) B.size(); i++)
                C[i+m] = (C[i+m]+c*B[i])%mod;
            L = n+1-L;
            B = T;
            b = d;
            m = 1;
        } else {
            ll c = mod-d*powmod(b,mod-2)%mod;
            while(SZ(C)<SZ(B)+m)
                C.push_back(0);
            for(int i=0; i<(int)B.size(); i++)
                C[i+m] = (C[i+m]+c*B[i])%mod;
            ++m;
        }
    }
    return C;
}

ll gao(VI a,ll n) {
    VI c = BM(a);
    c.erase(c.begin());
    for(int i=0; i<(int) c.size(); i++)
        c[i] = (mod-c[i])%mod;
    return (ll) solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}

}

int main() {
    ll t;
    ll nnn;
    VI a;
    a.push_back(3);
    a.push_back(9);
    a.push_back(20);
    a.push_back(46);
    a.push_back(106);
    a.push_back(244);
    a.push_back(560);
    a.push_back(1286);
    a.push_back(2956);
    a.push_back(6794);
    scanf("%lld",&t);
    while(t--) {
        scanf("%lld",&nnn);
        printf("%lld\n",linear_seq::gao(a,nnn-1));
    }
}

#include <bits/stdc++.h>

using namespace std;
int T;
int n,q;
int v,c;
typedef long long ll;
const ll mod=1e9+7;
ll dp[10005];
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,0,sizeof dp);
        dp[0]=1;
        scanf("%d %d",&n,&q);
        for(int i=1;i<=n;++i)
        {
            scanf("%d %d",&v,&c);
            for(int j=0;j<c;++j)
            {
                int V=v<<j;
                for(int k=10000;k>=V;--k)
                {
                    dp[k]+=dp[k-V];
                    dp[k]%=mod;
                }
            }
        }
        while(q--)
        {
            int x;
            scanf("%d",&x);
            printf("%lld\n",dp[x]);
        }
    }
    return 0;
}

import java.util.Scanner;
import java.math.BigInteger;
public class Main {
    public static void main(String[] args){
        Scanner cin=new Scanner(System.in);
        int T=cin.nextInt();
        for(int i=1;i<=T;i+=1){
            BigInteger S=cin.nextBigInteger();

            BigInteger k1=BigInteger.valueOf(1);

            BigInteger k2=BigInteger.valueOf(2);

            BigInteger k3=S.subtract(k1);

            BigInteger X=S.multiply(k3).divide(k2);

            BigInteger l=k1;
            BigInteger r=S;
            int f1=0;
            for (;;){
                BigInteger sum=l.add(r);
                BigInteger mid=sum.divide(k2);
                BigInteger k=mid.multiply(mid);
                int tmp=k.compareTo(S);
                if (tmp==0){
                    f1=1;
                    break;
                }
                if (tmp>0){
                    r=mid.subtract(k1);
                }
                if (tmp<0){
                    l=mid.add(k1);
                }
                int tmp1=l.compareTo(r);
                if (tmp1>0){
                    break;
                }

            }
            int f2=0;
            l=k1;
            r=X;
            for (;;){
                BigInteger sum=l.add(r);
                BigInteger mid=sum.divide(k2);
                BigInteger k=mid.multiply(mid);
                int tmp=k.compareTo(X);
                if (tmp==0){
                    f2=1;
                    break;
                }
                if (tmp>0){
                    r=mid.subtract(k1);
                }
                if (tmp<0){
                    l=mid.add(k1);
                }
                int tmp1=l.compareTo(r);
                if (tmp1>0){
                    break;
                }
            }

            if(f1==1&&f2==1){
                System.out.println("Arena of Valor");
            }
            if(f1==1&&f2==0){
                System.out.println("Hearth Stone");
            }
            if(f1==0&&f2==1){
                System.out.println("Clash Royale");
            }
            if(f1==0&&f2==0)
            {
                System.out.println("League of Legends");
            }
        }
    }
}

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
string str;
int a,b,c;
int main() {
    //freopen("1.txt","r",stdin);
    while(scanf("%d%d%d",&a,&b,&c)!=EOF){
        if(a%2==0||b%2==0||c%2==0){
            puts("Yes");
        }
        else puts("No");
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const ll mod=1000000007;
int T;
char str[maxn];
ll fast_pow(ll a,ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1)ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
int main() {
    //freopen("1.txt","r",stdin);
    scanf("%d", &T);
    while (T--) {
        scanf("%s", str);
        int len = strlen(str);
        ll cur = 0;
        for (int i = 0; i <len; i++) {
            cur = (cur * 10 + str[i] - ‘0‘) % (mod - 1);
        }
        cur = (cur - 1 + mod - 1) % (mod - 1);
        cur = fast_pow(2, cur);
        printf("%lld\n", cur);
    }
    return 0;
}

　

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
int T;
string str;
int main() {
    //freopen("1.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        cin>>str;
        int len=str.length();
        for (int i=0;i<len;i++){
            if (‘A‘<=str[i]&&str[i]<=‘Z‘){
                str[i]+=32;
            }
        }
        if (str=="jessie"){
            puts("Good guy!");
        }else{
            puts("Dare you say that again?");
        }
    }
    return 0;
}

　　

#include<bits/stdc++.h>

using namespace std;
typedef pair<int,int> PI;
const int maxn=5005;
const int M=200005;
const int inf=0x3f3f3f3f;
int a[M];
int n,k,m;
struct Point{
    int x,y,w;
}point[maxn];

struct Min_Cost_Max_Flow
{
    struct edge
    {
        int to,cap,cost,rev;
        edge() {};
        edge(int _to,int _cap,int _cost,int _rev):to(_to),cap(_cap),cost(_cost),rev(_rev) {};
    };
    vector<edge>E[maxn];
    int h[maxn],n,d[maxn],preV[maxn],preE[maxn];
    void init(int n)
    {
        this->n=n;
        for (int i=0; i<=n; i++)
        {
            E[i].clear();
            h[i]=0;
        }
    }
    void add(int from,int to,int cap,int cost)
    {
        E[from].push_back(edge(to,cap,cost,E[to].size()));
        E[to].push_back(edge(from,0,-cost,E[from].size()-1));
    }

    PI dijkstra(int s,int t,int f)
    {
        int cost=0,flow=0;
        for (int i=0; i<=n; i++)
        {
            h[i]=0;
        }
        while (f)
        {
            priority_queue<PI,vector<PI>,greater<PI> >q;
            for (int i=0; i<=n; i++)
            {
                d[i]=inf;
            }
            d[s]=0;
            q.push(make_pair(0,s));
            while (!q.empty())
            {
                PI now=q.top();
                q.pop();
                int v=now.second;
                if (d[v]<now.first)
                {
                    continue;
                }
                for (int i=0; i<E[v].size(); i++)
                {
                    edge &e=E[v][i];
                    if (e.cap>0&&d[e.to]>d[v]+e.cost+h[v]-h[e.to])
                    {
                        d[e.to]=d[v]+e.cost+h[v]-h[e.to];
                        preV[e.to]=v;
                        preE[e.to]=i;
                        q.push(make_pair(d[e.to],e.to));
                    }
                }
            }
            if (d[t]==inf)break;
            for (int i=0; i<=n; i++)
            {
                h[i]+=d[i];
            }
            int d=f;
            for (int i=t; i!=s; i=preV[i])
            {
                d=min(d,E[preV[i]][preE[i]].cap);
            }
            f-=d;
            flow+=d;
            cost+=d*h[t];
            for (int i=t; i!=s; i=preV[i])
            {
                edge &e=E[preV[i]][preE[i]];
                e.cap-=d;
                E[i][e.rev].cap+=d;
            }
        }
        return make_pair(flow,cost);
    }
} G;

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d%d",&n,&k,&m);
        for (int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].w);
            point[i].y++;
        }
        for (int i=1;i<=m;i++){
            a[i]=point[i].x;
            a[i+m]=point[i].y;
        }
        sort(a+1,a+m*2+1);
        int nn=unique(a+1,a+m*2+1)-a-1;
        G.init(nn+2);
        for (int i=1;i<=m;i++){
            point[i].x=lower_bound(a+1,a+1+nn,point[i].x)-a;
            point[i].y=lower_bound(a+1,a+1+nn,point[i].y)-a;
        }
        for (int i=1;i<=nn;i++) {
            G.add(i,i+1,k,0);
        }
        for (int i=1;i<=m;i++) {
            G.add(point[i].x,point[i].y,1,-point[i].w);
        }
        G.add(nn+2,1,k,0);
        PI ans=G.dijkstra(nn+2,nn+1,inf);
        printf("%d\n",-ans.second);
    }
}

原文地址：https://www.cnblogs.com/Accpted/p/11290610.html

时间： 2024-10-04 00:35:40

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题目链接:http://acm.fzu.edu.cn/contest/list.php?cid=158 A题题目: 题意: 给你六种操作:def, mul,mod,div, add, sub.除了看这几个字母也都知道是啥意思了,其中def是进行define. 思路: 比赛时队友写的,直接模拟,不过赛后补题时队友和我说mul时需要快速乘. 代码实现如下: 1 #include <set> 2 #include <map> 3 #include <queue> 4 #inc