1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 
 5 map<string,ll>mp;
 6 struct sair{
 7     string str;
 8     int id;
 9     ll num;
10 }a[1005];
11 
12 bool cmp(sair a,sair b){
13     if(a.num==b.num) return a.id<b.id;
14     return a.num>b.num;
15 }
16 
17 int main(){
18 
19     int n;
20     cin>>n;
21     string ans;
22     ll Max=-0x3f3f3f3f;
23     for(int i=1;i<=n;i++){
24         cin>>a[i].str>>a[i].num;
25         a[i].id=i;
26     }
27     for(int i=1;i<=n;i++){
28         mp[a[i].str]+=a[i].num;
29         a[i].num=mp[a[i].str];
30     }
31     for(int i=1;i<=n;i++){
32         if(mp[a[i].str]>Max){
33             Max=mp[a[i].str];
34         }
35     }
36     sort(a+1,a+n+1,cmp);
37     map<string,ll>tmp;
38     for(int i=1;i<=n;i++){
39         if(mp[a[i].str]==Max){
40             tmp[a[i].str]=1;
41         }
42     }
43     int idmin=0x3f3f3f3f; 
44     for(int i=1;i<=n;i++){
45         if(tmp[a[i].str]==1){
46             if(idmin>a[i].id&&a[i].num>=Max){
47                 idmin=a[i].id;
48                 ans=a[i].str;
49             }
50         }
51     }
52     cout<<ans<<endl;
53     //system("pause");
54 
55 }

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define lson l,mid,rt<<1
  4 #define rson mid+1,r,rt<<1|1
  5 typedef long long ll;
  6 
  7 int n;
  8 int dp[1005][1005][2];///dp[i][j][0] 表示2的数量，dp[i][j][1] 表示5的数量
  9 int pre[1005][1005][2];
 10 
 11 void print(int i,int j,int k,int flag){
 12     if(i==1&&j==1);
 13     else if(i==1) print(i,j-1,k,0);
 14     else if(j==1) print(i-1,j,k,1);
 15     else{
 16         if(dp[i][j][k]==dp[i][j-1][k]+pre[i][j][k])
 17             print(i,j-1,k,0);
 18         else
 19             print(i-1,j,k,1);
 20     }
 21     if(flag==2) return;
 22     cout<<(flag?"D":"R");
 23 }
 24 
 25 int main(){
 26     #ifndef ONLINE_JUDGE
 27         freopen("1.txt","r",stdin);
 28     #endif
 29     std::ios::sync_with_stdio(false);
 30     cin>>n;
 31     int num,tmp;
 32     int x=-1,y=-1;
 33     memset(dp,0x3f,sizeof(dp));
 34     for(int i=1;i<=n;i++){
 35         for(int j=1;j<=n;j++){
 36             cin>>num;
 37             if(!num){
 38                 pre[i][j][0]++;
 39                 pre[i][j][1]++;
 40                 x=i,y=j;
 41                 continue;
 42             }
 43             tmp=num;
 44             while(tmp%2==0){
 45                 pre[i][j][0]++;
 46                 tmp/=2;
 47             }
 48             tmp=num;
 49             while(tmp%5==0){
 50                 pre[i][j][1]++;
 51                 tmp/=5;
 52             }
 53         }
 54     }
 55 
 56     for(int i=1;i<=n;i++){
 57         for(int j=1;j<=n;j++){
 58             for(int k=0;k<2;k++){
 59                 if(i>1){
 60                     dp[i][j][k]=min(dp[i][j][k],dp[i-1][j][k]);
 61                 }
 62                 if(j>1){
 63                     dp[i][j][k]=min(dp[i][j][k],dp[i][j-1][k]);
 64                 }
 65                 if( i==1 && j==1 ){
 66                     dp[i][j][k]=0;
 67                 }
 68                 dp[i][j][k]+=pre[i][j][k];
 69             }
 70         }
 71     }
 72     int ans=min(dp[n][n][0],dp[n][n][1]);
 73     if(x!=-1&&y!=-1&&ans>=1){
 74         cout<<1<<endl;
 75         for(int i=1;i<x;i++){
 76             cout<<"D";
 77         }
 78         for(int i=1;i<y;i++){
 79             cout<<"R";
 80         }
 81         for(int i=x;i<n;i++){
 82             cout<<"D";
 83         }
 84         for(int i=y;i<n;i++){
 85             cout<<"R";
 86         }
 87         cout<<endl;
 88     }
 89     else{
 90         cout<<ans<<endl;
 91         if(dp[n][n][0]>dp[n][n][1]){
 92             print(n,n,1,2);
 93         }
 94         else{
 95             print(n,n,0,2);
 96         }
 97         cout<<endl;
 98     }
 99     return 0;
100 }

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 typedef long long ll;
 7 
 8 struct circle{
 9     double x,y,r;
10 }c[5];
11 
12 double angle[5];
13 
14 double Check(double x,double y){
15     for(int i=0;i<3;i++){
16         angle[i]=sqrt(sqr(x-c[i].x)+sqr(y-c[i].y))/c[i].r;
17     }
18     double ans=0;
19     for(int i=0;i<3;i++){
20         ans+=sqr(angle[i]-angle[(i+1)%3]);
21     }
22     return ans;
23 }
24 
25 int main(){
26     #ifndef ONLINE_JUDGE
27         freopen("1.txt","r",stdin);
28     #endif
29     for(int i=0;i<3;i++){
30         scanf("%lf %lf %lf",&c[i].x,&c[i].y,&c[i].r);
31     }
32     double ansx,ansy;
33     ansx=(c[0].x+c[1].x+c[2].x)/3;
34     ansy=(c[0].y+c[1].y+c[2].y)/3;
35     double step=1.0;
36     int flag;
37     double tmp;
38     while(step>1e-5){
39         flag=0;
40         tmp=Check(ansx,ansy);
41         if(tmp>Check(ansx+step,ansy)) ansx+=step,flag=1;
42         else if(tmp>Check(ansx-step,ansy)) ansx-=step,flag=1;
43         else if(tmp>Check(ansx,ansy+step)) ansy+=step,flag=1;
44         else if(tmp>Check(ansx,ansy-step)) ansy-=step,flag=1;
45         if(!flag) step*=0.8;
46     }
47     if(Check(ansx,ansy)<1e-5) printf("%.6f %.6f\n",ansx,ansy);
48 
49 }