2018.3.25校内互测

思路：模拟搞一下就出来了。

#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 1010
using namespace std;
char s[101],ss[1001];
int top1,top2,num,now;
int stack1[MAXN],stack2[MAXN];
map<int,string>ma;
map<string,int>mm;
int main(){
    //freopen("lppin.txt","r",stdin);
    freopen("kami.in","r",stdin);
    freopen("kami.out","w",stdout);
    ma[++num]="http://www.acm.org/";
    mm["http://www.acm.org/"]=num;
    now=1;
    while(cin>>s&&s[0]!='Q'){
        if(s[0]=='V'){
            cin>>ss;
            if(!mm[ss]){
                ma[++num]=ss;
                mm[ss]=num;
            }
            top1++;stack1[top1]=now;
            now=mm[ss];
            cout<<ss<<endl;top2=0;
        }
        else if(s[0]=='B'){
            if(top1){
                top2++;stack2[top2]=now;
                now=stack1[top1];top1--;
                cout<<ma[now]<<endl;
            }    
            else cout<<"Ignored"<<endl;
        }
        else if(s[0]=='F'){
            if(top2){
                top1++;stack1[top1]=now;
                now=stack2[top2];top2--;
                cout<<ma[now]<<endl;
            }
            else cout<<"Ignored"<<endl;
        }
    }
}
/*
VISIT http://acm.ashland.edu/
VISIT http://acm.baylor.edu/acmicpc/
BACK
BACK
BACK
FORWARD
VISIT http://www.ibm.com/
BACK
BACK
FORWARD
FORWARD
FORWARD
QUIT
*/

 

思路：40分矩阵快速幂暴力

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mod 1000000007
using namespace std;
typedef long long LL;
int T,n;
LL t[2][2],ans[2][2],r[2][2];
void mul(LL a[2][2],LL b[2][2]){
    memset(r,0,sizeof(r));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                r[i][j]+=a[i][k]*b[k][j],r[i][j]%=mod;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            a[i][j]=r[i][j];
}
void mul1(LL a[2][2],LL b[2][2]){
    memset(r,0,sizeof(r));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                r[i][j]+=a[i][k]*b[k][j];
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            a[i][j]=r[i][j];
}
int main(){
    //freopen("lppin.txt","r",stdin);
    freopen("na.in","r",stdin);
    freopen("na.out","w",stdout);
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        memset(ans,0,sizeof(ans));
        memset(t,0,sizeof(t));
        if(n==0){ cout<<"0"<<endl; continue; }
        t[0][0]=t[0][1]=t[1][0]=1;
        ans[0][0]=ans[0][1]=1;
        long long p;
        if(n==1||n==2)    p=1;
        else{
            n-=2;
            while(n){
                if(n&1) mul1(ans,t);
                mul1(t,t); n>>=1;
            }
        } 
        p=ans[0][0];
        memset(ans,0,sizeof(ans));
        memset(t,0,sizeof(t));
        t[0][0]=t[0][1]=t[1][0]=1;
        ans[0][0]=ans[0][1]=1;
        if(p==1||p==2){ cout<<"1"<<endl; continue; }
        else{
            p-=2;
            while(p){
                if(p&1) mul(ans,t);
                mul(t,t); p>>=1;
            }
        }
        cout<<ans[0][0]<<endl;
    }
}

题解：（╮(╯▽╰)╭虽然有几个地方并没有看懂）

#include<cstdio>
#include<cstdlib>
#include<cstring>

using namespace std;

#ifdef unix
#define LL "%lld"
#else
#define LL "%I64d"
#endif

const int maxn=200;
const long long mo=1000000007;
const long long mo2=mo*2+2;
const long long mo3=mo2*3;

char s[maxn];

struct matrix
{
    long long z[3][3];
    matrix()
    {
        memset(z,0,sizeof(z));
    }
    matrix operator*(const matrix &a)const
    {
        matrix ans;
        for (int b=1;b<=2;b++)
            for (int c=1;c<=2;c++)
                for (int d=1;d<=2;d++)
                    ans.z[b][c]=(ans.z[b][c]+z[b][d]*a.z[d][c]%mo)%mo;
        return ans;
    }
    matrix operator+(const matrix &a)const
    {
        matrix ans;
        for (int b=1;b<=2;b++)
            for (int c=1;c<=2;c++)
                for (int d=1;d<=2;d++)
                    ans.z[b][c]=(ans.z[b][c]+z[b][d]*a.z[d][c]%mo2)%mo2;
        return ans;
    }
}m1,m2;

struct bign
{
    int z[maxn],l;
    void init()
    {
        memset(z,0,sizeof(z));
        scanf("%s",s+1);
        l=strlen(s+1);
        for (int a=1;a<=l;a++)
            z[a]=s[l-a+1]-'0';
    }
    long long operator%(const long long &a)const
    {
        long long b=0;
        for (int c=l;c>=1;c--)
            b=(b*10+z[c])%a;
        return b;
    }
}z;

long long get(long long v)
{
    if (v==0) return 0;
    m1.z[1][1]=0;
    m1.z[1][2]=1;
    m2.z[1][1]=0;
    m2.z[1][2]=m2.z[2][1]=m2.z[2][2]=1;
    while (v)
    {
        if (v&1) m1=m1*m2;
        m2=m2*m2;
        v>>=1;
    }
    return m1.z[1][1];
}

long long get1(long long v)
{
    if (v==0) return 0;
    m1.z[1][1]=0;
    m1.z[1][2]=1;
    m2.z[1][1]=0;
    m2.z[1][2]=m2.z[2][1]=m2.z[2][2]=1;
    while (v)
    {
        if (v&1) m1=m1+m2;
        m2=m2+m2;
        v>>=1;
    }
    return m1.z[1][1];
}

int main()
{
    freopen("na.in","r",stdin);
    freopen("na.out","w",stdout);

    int t;
    scanf("%d",&t);
    for (int a=1;a<=t;a++)
    {
        z.init();
        long long v1=z % mo3;
        v1=get1(v1);
        printf(LL "
",get(v1));
    }

    return 0;
}

思路：

想写30分的暴力，还挂了一个点，30分暴力：

#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 510
using namespace std;
map<string,int>ma;
int vis[MAXN];
int n,H,mmp,ansval=-1,anscost=0x7f7f7f7f,ans;
struct nond{
    int job,val,cost;
}v[MAXN];
void dfs(int now,int sum,int val,int cost,int door,int back,int mid,int front,int head){
    if(sum+n-now+1<11)    return ;
    if(cost>H)    return ;
    if(sum==11&&door!=0&&back>=3&&back<=5&&mid>=2&&mid<=5&&front>=1&&front<=3&&head!=0){
        string s;
        for(int i=1;i<=n;i++)    s+=char(vis[i]+'0');
        s+=char(head+'0');
        if(val>ansval){
            ansval=val;anscost=cost;
            ans=1;ma[s]=1;
        }
        else if(val==ansval){
            if(anscost==cost){
                if(!ma[s]){
                    if(val==716&&ans==1){
                        int hh=1;
                    }
                    ma[s]=1;ans++;
                }
            }
            else if(cost<anscost){ anscost=cost; ans=1; }
        }
        return ;
    }
    if(now>n)    return ;
    if(v[now].job==1){
        if(door==0)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val,cost+v[now].cost,1,back,mid,front,head),vis[now]=0;
        if(door==0&&head==0)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val*2,cost+v[now].cost,1,back,mid,front,now),vis[now]=0;
        dfs(now+1,sum,val,cost,door,back,mid,front,head);    
    }
    else if(v[now].job==2){
        if(back<5)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val,cost+v[now].cost,door,back+1,mid,front,head),vis[now]=0;
        if(back<5&&head==0)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val*2,cost+v[now].cost,door,back+1,mid,front,now),vis[now]=0;
        dfs(now+1,sum,val,cost,door,back,mid,front,head);
    }
    else if(v[now].job==3){
        if(mid<5)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val,cost+v[now].cost,door,back,mid+1,front,head),vis[now]=0;
        if(mid<5&&head==0)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val*2,cost+v[now].cost,door,back,mid+1,front,now),vis[now]=0;
        dfs(now+1,sum,val,cost,door,back,mid,front,head);
    }
    else if(v[now].job==4){
        if(front<3)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val,cost+v[now].cost,door,back,mid,front+1,head),vis[now]=0;
        if(front<3&&head==0)    vis[now]=1,dfs(now+1,sum+1,val+v[now].val*2,cost+v[now].cost,door,back,mid,front+1,now),vis[now]=0;
        dfs(now+1,sum,val,cost,door,back,mid,front,head);
    }
}
int cmp(nond a,nond b){
    if(a.val==b.val)    return a.cost<b.cost;
    return a.val>b.val;
}
int main(){
    //freopen("lppin.txt","r",stdin);
    freopen("wosa1.in","r",stdin);
    freopen("wosa.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        char c[100];cin>>c;
        int y,z;cin>>y>>z;mmp+=z;
        if(c[0]=='G')    v[i].job=1;
        else if(c[0]=='D')    v[i].job=2;
        else if(c[0]=='M')    v[i].job=3;
        else if(c[0]=='F')    v[i].job=4;
        v[i].val=y;v[i].cost=z;
    }
    cin>>H;
    if(n<=20){
        dfs(1,0,0,0,0,0,0,0,0);
        cout<<ansval<<" "<<anscost<<" "<<ans;
    }
    else if(H>=mmp){
        sort(v+1,v+1+n,cmp);
        for(int a=3;a<=5;a++)
            for(int b=2;b<=5;b++){
                int c=11-1-a-b;
                int door=0,back=0,mid=0,front=0,val=0,cost=0;
                if(c<1)    continue;
                for(int i=1;i<=n;i++){
                    if(v[i].job==1){
                        if(door==0)    door=1,val+=v[i].val,cost+=v[i].cost;
                    }
                    else if(v[i].job==2){
                        if(back<a)    back++,val+=v[i].val,cost+=v[i].cost;
                    }
                    else if(v[i].job==3){
                        if(mid<b)    mid++,val+=v[i].val,cost+=v[i].cost;
                    }
                    else if(v[i].job==4){
                        if(front<c)    front++,val+=v[i].val,cost+=v[i].cost;
                    }
                }
                val+=v[1].val;
                if(val>ansval){ ansval=val; ans=1; }
                if(val==ansval){
                    if(cost==anscost)    ans++;
                    else if(cost<anscost){ anscost=cost;ans=1; }
                }
            }
        if(ans>100)    ans=1000000000;
        cout<<ansval<<" "<<anscost<<" "<<ans;
    }
    else {
        dfs(1,0,0,0,0,0,0,0,0);
        cout<<ansval<<" "<<anscost<<" "<<ans;
    }
}
/*
15
Defender 23 45
Midfielder 178 85
Goalkeeper 57 50
Goalkeeper 57 50
Defender 0 45
Forward 6 60
Midfielder 20 50
Goalkeeper 0 50
Midfielder 64 65
Midfielder 109 70
Forward 211 100
Defender 0 40
Defender 29 45
Midfielder 57 60
Defender 52 45
600
*/

题解：见代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mod 1000000000
using namespace std;
struct nond{
    int job,val,cost;    
}v[501];
int n,H,ansval=-1,anscost,ans;
int L[5],R[5];
int f[2][6][6][4][1001][3];
/*f[a][b][c][d][e][f]表示到花费了e元为至
 a表示选的守门员，b表示选了几个后卫，c表示选了几个中锋，
 e表示选了几个前锋
 f如果等于0表示价值，如果是1表示方案数，如果是2表示队长价值*/
int cmp(nond a,nond b){
    if(a.val==b.val)    return a.cost>b.cost;
    return a.val<b.val;
}
void up(int a,int b,int c,int d,int e,int val,int num,int mx){
    if(f[a][b][c][d][e][0]<val){
        f[a][b][c][d][e][0]=val;
        f[a][b][c][d][e][1]=num;
        f[a][b][c][d][e][2]=mx;
    }
    else if(f[a][b][c][d][e][0]==val&&f[a][b][c][d][e][2]<mx){
        f[a][b][c][d][e][1]=num; 
        f[a][b][c][d][e][2]=mx;
    }
    else if(f[a][b][c][d][e][0]==val&&f[a][b][c][d][e][2]==mx){
        f[a][b][c][d][e][1]+=num;
        if(f[a][b][c][d][e][1]>mod)
            f[a][b][c][d][e][1]=mod;
    }
}
/*背包*/
void insert(int x){
    for(int a=R[1]-(v[x].job==1);a>=0;a--)
        for(int b=R[2]-(v[x].job==2);b>=0;b--)
            for(int c=R[3]-(v[x].job==3);c>=0;c--)
                for(int d=R[4]-(v[x].job==4);d>=0;d--)
                    for(int e=H-v[x].cost;e>=0;e--)
                        if(f[a][b][c][d][e][1]){
                            int val=v[x].val+f[a][b][c][d][e][0];
                            up(a+(v[x].job==1),b+(v[x].job==2),c+(v[x].job==3),d+(v[x].job==4),e+v[x].cost,val,f[a][b][c][d][e][1],v[x].val); 
                        }
} 
int main(){
    freopen("wosa4.in","r",stdin);
    //freopen("wosa.out","w",stdout);
    //freopen("1wosaout.txt","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        char s[100];cin>>s;
        int x,y;cin>>x>>y;
        if(s[0]=='G')    v[i].job=1;
        if(s[0]=='D')    v[i].job=2;
        if(s[0]=='M')    v[i].job=3;
        if(s[0]=='F')    v[i].job=4;
        v[i].val=x;v[i].cost=y;
    }
    scanf("%d",&H);
    L[1]=1;R[1]=1;L[2]=3;R[2]=5;
    L[3]=2;R[3]=5;L[4]=1;R[4]=3;
    //分别表示4种人所选的人数的限制范围。 
    sort(v+1,v+1+n,cmp);//预处理排序（不明白啊啊啊啊，为什么这么排） 
    f[0][0][0][0][0][1]=1;    //没选的方案数。 
    f[0][0][0][0][0][2]=-1;    //没选的队长价值（没有队长，所以设成-1） 
    for(int i=1;i<=n;i++)    insert(i);
    for(int a=L[1];a<=R[1];a++)
        for(int b=L[2];b<=R[2];b++)
            for(int c=L[3];c<=R[3];c++)
                for(int d=L[4];d<=R[4];d++)
                    if(a+b+c+d==11)
                        for(int e=0;e<=H;e++){
                            if(f[a][b][c][d][e][1]==0)    continue;
                            int val=f[a][b][c][d][e][0]+f[a][b][c][d][e][2];
                            if(val>ansval){
                                ansval=val;anscost=e;
                                ans=f[a][b][c][d][e][1];
                            }
                            else if(val==ansval&&e<anscost){
                                anscost=e;
                                ans=f[a][b][c][d][e][1];
                            }
                            else if(val==ansval&&e==anscost){
                                ans+=f[a][b][c][d][e][1];
                                if(ans>mod)    ans=mod;
                            }
                        }
    cout<<ansval<<" "<<anscost<<" "<<ans;
}
/*
15
Defender 23 45
Midfielder 178 85
Goalkeeper 57 50
Goalkeeper 57 50
Defender 0 45
Forward 6 60
Midfielder 20 50
Goalkeeper 0 50
Midfielder 64 65
Midfielder 109 70
Forward 211 100
Defender 0 40
Defender 29 45
Midfielder 57 60
Defender 52 45
600
*/