Contest1036

一名大三older 希望能帮助大一萌新。。。加油 xiandianer!!!!;

xdoj 1260 (B) 水题 用了一下STL (好长啊。。。）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
const int N=107;
struct T{
    string suj;
    double  p;
    bool operator <(const T& b) const {
        return suj<b.suj;
    }
};
map <T,int> mapp;
int n;
double f (int x) {
    if (x<=59) return 0.0;
    if (x<=62) return 1.0;
    if (x<=65) return 1.5;
    if (x<=68) return 1.7;
    if (x<=71) return 2.0;
    if (x<=74) return 2.3;
    if (x<=77) return 2.7;
    if (x<=81) return 3.0;
    if (x<=84) return 3.3;
    if (x<=89) return 3.7;
    return 4.0;
}
int main ()
{
    cin>>n;
    map<T,int>::iterator it;
    for (int i=1;i<=n;i++) {
        T temp;
        int w;
        cin>>temp.suj>>temp.p>>w;
        if (mapp[temp]<w)
            mapp[temp]=w;
    }
    it=mapp.begin();
    double  sum=0.0;
    double  k=0.0;
    double  nn=0.0;
    while (it!=mapp.end()) {
        sum+=(it->first.p)*(it->second);
        k+=(it->first.p)*f(it->second);
        nn+=it->first.p;
        it++;
    }
    printf ("%.1lf
%.2lf
",sum/nn,k/nn);
    return 0;
}

xdoj 1255 (C) 水题 不明白为什么不加getchar() 会GG;

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
char* str="0123456789ABCDEF";
int main ()
{
    int x,y,z;
    while (~scanf("(%d, %d, %d)",&x,&y,&z)) {
           char s[7];
           s[0]=str[x/16];
           s[1]=str[x%16];
           s[2]=str[y/16];
           s[3]=str[y%16];
           s[4]=str[z/16];
           s[5]=str[z%16];
           s[6]='';
           cout<<"#"<<s<<endl;
           getchar();
    }
    return 0;
}

xdoj 1265(D） 取余找循环节呗。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1010;
int mod=1001;
struct T {
    int a;
    int b;
};
T t[N];
int p[N];
int A,B,C;
int n,k;
int mycmp (T aa,T bb) {
    return aa.a>bb.a;
}
int main ()
{
    int nn;
    scanf ("%d",&nn);
    while (nn--) {
        scanf ("%d %d %d",&A,&B,&C);
        memset (p,0,sizeof(p));
        int i=1;
        t[1].a=A;
        p[A]=1;
        while (1) {
            t[i+1].a=(t[i].a*B+C)%mod;
            if (p[t[i+1].a])
                break;
            else
                p[t[i+1].a]=i+1;
            i++;
        }
        int x1=p[t[i+1].a];
        int x2=i;
        int num=x2-x1+1;
        scanf ("%d %d",&n,&k);
        int times=(n-x1+1)/num;
        for (i=1;i<=x2;i++) {
            if (i<x1) t[i].b=1;
            else if (i<=x1+(n-x1+1)%num)
                t[i].b=times+1;
            else
                t[i].b=times;
        }
        sort (t+1,t+1+x2,mycmp);
        i=1;
        while (1) {
            if (p[t[i].a]>n) {i++; continue;}
            if (k-t[i].b<=0) break;
            else             k-=t[i].b;
            i++;
        }
        printf ("%d
",t[i].a);
    }
    return 0;
}

xdoj 1262（E） 找A前后Q数目相乘

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N=1e5+7;
char str[N];
int main ()
{
    scanf ("%s",str);
    int len=strlen(str);
    LL n1=0;
    for (int i=0;i<len;i++) {
        if (str[i]=='Q')
            n1++;
    }
    LL n2=0;
    LL sum=0;
    for (int i=0;i<len;i++) {
        if (str[i]=='Q')
            n2++;
        if (str[i]=='A')
            sum+=n2*(n1-n2);
    }
    printf ("%lld
",sum);
    return 0;
}

xdoj  1264(F) 找一下规律 发现数据按照二进制递增的 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long LL;
bool a[107];
int main ()
{
    LL n;
    while (~scanf ("%lld",&n)) {
        memset (a,0,sizeof(a));
        int i=0;
        while (n) {
            if (n&1) a[i]=1;
            n=n>>1;
            i++;
        }
        for (int i=99;i>=0;i--) {
            if (a[i])
                printf ("R");
            else
                printf ("G");
        }
        printf ("
");
    }
    return 0;
}

xdoj 1261(G)  开始用图做的 GG了最后用的并查集

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=1e6+7;
int p[N];
int f[N];
int n,m;
int _find(int x) {
    if (x!=f[x])
        f[x]=_find(f[x]);
    return f[x];
}
int main ()
{
    while (~scanf ("%d %d",&n,&m)) {
        for (int i=1;i<=n;i++)
            f[i]=i;
        memset(p,0,sizeof(p));
        for (int i=1;i<=m;i++) {
            int op,a,b;
            scanf ("%d %d %d",&op,&a,&b);
            if (op==0) {
                int x1=_find(a);
                int x2=_find(b);
                if (x1==x2||p[x1]==x2) continue;
                if (!p[x1]&&!p[x2]) {
                    p[x1]=x2;
                    p[x2]=x1;
                }
                else if (!p[x1]&&p[x2]) f[x1]=p[x2];
                else if (!p[x2]&&p[x1]) f[x2]=p[x1];
                else    {f[x2]=p[x1];   f[p[x2]]=x1;}
                //for (int i=1;i<=n;i++)
                   // cout<<_find(i)<<"  "<<p[i]<<endl;
            }
            else {
                int x1=_find(a);
                int x2=_find(b);
                if (x1==x2)
                    printf ("In the same category.
");
                else if (p[x1]==x2)
                    printf ("In different category.
");
                else
                    printf ("Not sure yet.
");
            }
        }
    }
    return 0;
}

xdoj  1254(H) 大整数减法 基本功

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int N=107;
int a[N];
int b[N];
int x1[N];
int x2[N];
string s1,s2;
int  cmp (int a[],int b[]) {
    for (int i=101;i>=0;i--) {
        if (a[i]>b[i]) return 1;
        if (a[i]<b[i]) return -1;
    }
    return 0;
}
void f (int x[]) {
    memset(x1,0,sizeof(x1));
    memset(x2,0,sizeof(x2));
    int j=0;
    for (int i=s1.size()-1;i>=0;i--)
        x1[j++]=s1[i]-'0';
    j=0;
    for (int i=s2.size()-1;i>=0;i--)
        x2[j++]=s2[i]-'0';
    for (int i=0;i<s1.size();i++) {
        if (x1[i]<x2[i]) {
            j=i;
            while (x1[j+1]==0) {
                  x1[j+1]=9;
                  j++;
            }
        x1[j+1]--;
        x1[i]+=10;
      }
    x[i]=(x1[i]-=x2[i]);
   }
   return ;
}
int main ()
{
    cin>>s2>>s1;
    f(a);
    cin>>s2>>s1;
    f(b);
    if (cmp(a,b)>0)
        printf ("Ting
");
    else if (cmp(a,b)==0)
        printf ("Excellent
");
    else
        printf ("Yu
");
    return 0;
}

xdoj 1257 (I) 曼哈顿距离 我也是百度的。。。哈哈哈

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
int T,n;
ll a1,a2,b1,b2;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        a1=b1=INF,a2=b2=-INF;
        for(int i=0;i<n;i++)
        {
            ll x,y;
            scanf ("%lld %lld",&x,&y);
            a1=min(a1,x+y),a2=max(a2,x+y);
            b1=min(b1,x-y),b2=max(b2,x-y);
        }
        ll ans=max(a2-a1,b2-b1);
        printf("%lld
",ans);
    }
    return 0;
}

xdoj 1250 (J) 好吧这道题我想了好久。。一天？/ 最后感谢formilk的i提示 用的先序遍历 。。居然过了。但应该不是最好的解法

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=5e6+7;
struct T {
    int x;
    int y;
 };
T k[100];
T s[2*N];
int ss;
int t;
int n_max;
int a,b,c,d,n,m;
void fare (int a,int b,int c,int d) {
    if (b+d>n_max) return ;
    fare (a,b,a+c,b+d);
    k[++t].x=(a+c);
    k[t].y=(b+d);
    fare (a+c,b+d,c,d);
    return ;
}
int gcd (int x,int y) {
    return !y?x:gcd(y,x%y);
}
void f(int& x,int& y ) {
    int p=gcd (x,y);
    x=x/p;
    y=y/p;
    return ;
}
int main ()
{
    int Q;
    scanf ("%d",&Q);
    while (Q--) {
        scanf ("%d %d %d %d %d %d",&a,&b,&c,&d,&n,&m);
        ss=0;
        f(a,b); T t1={a,b};
        f(c,d); T t2={c,d};
        n_max=max(b,d);
        t=1;
        k[t].x=0; k[t].y=1;
        fare (0,1,1,1);
        k[++t].x=1;k[t].y=1;
        int i;
        for (i=1;;i++) if (k[i].x==t2.x&&k[i].y==t2.y) break;
        for (;;i--)  {
            s[++ss]=k[i];
            if (k[i].x==t1.x&&k[i].y==t1.y) break;
        }
        int nn=0;
        while (1){
           t1=s[ss--];
           nn++;
           if (nn==n+1) break;
           if (ss==0) break;
           t2=s[ss];
           while (1) {
                int x=t1.x+t2.x;
                int y=t1.y+t2.y;
                f(x,y);
                if (y<=m) {
                 t2.x=x; t2.y=y;
                 s[++ss]=t2;
                }
                else
                    break;
           }
        }
        if (nn==n+1&&!(t1.x==c&&t1.y==d))
            printf ("%d %d
",t1.x,t1.y);
        else
            printf ("-1
");
    }
    return 0;
}

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