Test 1 for NOIP: Result for Day2

头大

这个暑假完就要去搞NOIP了。。。

暑假55天也就20次测试。。。为防万一我还是每次测试玩都写个总结。。

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Day2 （120/300）

T1 能源 (100/100)

题目

简单二分即可a_a

MY.CPP/STD.CPP

#include<iostream>#include<iomanip>#include<cstdlib>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<string>#include<cmath>using namespace std;int n,ai,sum,maxx,power[1005],bater[1005];double k,ans;void find(){    double l=1,r=maxx;    while(r-l >= 0.00000001)    {        int jud = (l+r)/2;;        double mid = (l+r)/2;        double index = k/100;        double trie = ((power[1000]-power[jud])-mid*(bater[1000]-bater[jud]))*(1-index);        double limt = (bater[jud]*mid-power[jud]);        if(trie<=limt)  r = mid;        if(trie>limt)   l = mid;    }    printf("%0.6lf",l);}int  main(){    cin >> n >> k;    for(int i=1;i<=n;i++)    {        cin >> ai;        power[ai] += ai;        bater[ai] += 1;        maxx = max(maxx,ai);        sum += ai;    }    for(int i=1;i<=1000;i++)    {        power[i] += power[i-1];        bater[i] += bater[i-1];    }    find();}

T2 电影 (0/100)

题目

一开始我以为要用数位dp就没做，后来想水几分结果答案没有-1和w的点。。。
行吧我认输

STD.CPP

#include<iostream>#include<iomanip>#include<cstdio>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<map>#include<string>#include<cstring>using namespace std;int n,m,k,d,e,h;long long a,l,ans,i,j;double b[200005],c[200005];map<double,int> f;map<double,int> g;int gcd(long long a,long long b){    if(!a)  return b;    return gcd(b%a,a);}inline long long fan(long long x){    long long k=x,a=0;    for(;k;k/=10)        a = a*10+k%10;    return a;}int main(){    cin >> n >> m >> k;    if(n>m) swap(n,m);    ans = (n+1)*(m+1);    for(i=1;i<=max(n,m);i++)    {        a = fan(i);        l = gcd(a,i);        b[i] = (double)(a/l)/(i/l);        c[i] = (double)(i/l)/(a/l);    }    for(i=1;i<=n;i++)f[b[i]]++;    for(j=0;j<m&&h<k;)    {        j++;    g[b[j]]++;        h += f[c[j]];    }    if(h<k)    {cout << -1 << endl;return 0;}    i = n;  ans = i*j;    for(;i;i--)    {        f[b[i]]--;  h -= g[c[i]];        for(;j<m&&h<k;)         {            j++;    h += f[c[j]];            g[b[j]]++;        }        if(h<k) break;        ans = min(ans,(i-1)*j);    }    cout << ans;}

话说map好玩，vector也是

T3 流（网络流） (20/100)

题目

MY.CPP

#include<iostream>#include<iomanip>#include<cstdlib>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<string>#include<cmath>using namespace std;int n,inmini,inmaxx,otmini,otmaxx;bool exist[20];int check[10],value;struct node1{    int u;    int v;    int minn;    int maxx;    int val;}side[20];struct node2{    int in;    int ot;    int inmin;  int inmax;    int otmin;  int otmax;}dot[10];void dfs(int cn,int n){    exist[cn] = true;    for(int i=1;i<=(n*(n-1)/2);i++)    {        int h = side[i].v;        if(side[i].u==cn&&!exist[side[i].v])        {            int hihi1,hihi2;            hihi1 = max(dot[h].inmin,dot[h].otmin);            hihi2 = min(dot[h].inmax,dot[h].otmax);            check[h] += hihi1;            value += hihi2*hihi2 + side[i].val;            dfs(h,n);        }       }}int main(){    freopen("flow.in","r",stdin);    freopen("flow.out","w",stdout);    int n;    cin >> n;    for(int i=1;i<=(n*(n-1)/2);i++)    {        cin >> side[i].u >> side[i].v >> side[i].minn >> side[i].maxx >> side[i].val;        dot[side[i].u].ot +=1;      dot[side[i].v].in +=1;        dot[side[i].u].otmin += side[i].minn;        dot[side[i].u].otmax += side[i].maxx;               dot[side[i].v].inmin += side[i].minn;        dot[side[i].v].inmax += side[i].maxx;    }    if(n==2)        {        cout << side[1].minn << " " << side[1].minn*side[1].minn+side[1].val;        return 0;    }    for(int i=2;i<=n-1;i++)    {        if(dot[i].inmax<dot[i].otmin||dot[i].inmin>dot[i].otmax)        {            cout << "-1 -1" << endl;            return 0;        }    }    dfs(1,n);    cout << check[n] << " " << value;}

过了两个点但完全没有参考价值思路是完全错的。。
神奇的是样例数据全过
网络流我还是渣。。。

STD.CPP（玄学的网络流）

#include <cstdio>#include <algorithm>using namespace std;int a,b,c,d,e,f[10][10],in[10],out[10],g,h,i,j,k,l,m,n,Min[10][10],Max[10][10],sum[10][10],ansmax,ansflow;inline void dfs(int x,int y,int flow,int ans){    if(flow>ansflow)return;    if(x==n){        l=1;        if(flow<ansflow){ansflow=flow;ansmax=ans;}        else if(flow==ansflow){ansmax=max(ans,ansmax);}        return;    }    if(y>n){dfs(x+1,x+2,flow,ans);return;}    if(x>1&&y==n){        int s=in[x];        s-=out[x];        if(s>Max[x][y]||s<Min[x][y])return;        dfs(x,y+1,flow,ans+s*s+(s>0)*(sum[x][y]));        return ;    }    for(int i=Min[x][y];i<=Max[x][y];i++){        out[x]+=i;in[y]+=i;        if(x==1)dfs(x,y+1,flow+i,ans+i*i+(i>0)*sum[x][y]);        else dfs(x,y+1,flow,ans+i*i+(i>0)*sum[x][y]);        out[x]-=i;in[y]-=i;    }}int main(){    freopen("flow.in","r",stdin);    freopen("flow.out","w",stdout);    scanf("%d",&n);    if(n==1){        printf("%d %d",0,0);    }    else if(n==2){        scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);        if(c)printf("%d %d",c,e+c*c);        else printf("%d %d",c,0);    }    else if(n==3){    for(i=1;i<=3;i++){        scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);        f[a][b]=1;        Min[a][b]=c;        Max[a][b]=d;        sum[a][b]=e;    }    if(Max[1][2]<Min[2][3]||Min[1][2]>Max[2][3]){printf("-1 -1");return 0;}    h=max(Min[1][2],Min[2][3]);if(h)g=sum[1][2]+h*h+sum[2][3]+h*h;    h+=Min[1][3];if(Min[1][3])g+=sum[1][3]+Min[1][3]*Min[1][3];    printf("%d %d",h,g);    }    else{        //printf("-1 -1");return 0;        m=n*(n-1)/2;ansflow=10000000;        for(i=1;i<=m;i++){            scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);            f[a][b]=1;            Min[a][b]=c;            Max[a][b]=d;            sum[a][b]=e;        }        dfs(1,2,0,0);        if(l)printf("%d %d",ansflow,ansmax);        else printf("-1 -1");    }    return 0;}}