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  • 山东省第八届ACM大学生程序设计竞赛

    C 暂时还不对

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cmath>
     4 
     5 using namespace std;
     6 
     7 #define MOD 1000000007
     8 typedef long long ll;
     9 
    10 int jie[100010];
    11 
    12 void jiecheng1(){
    13     ll jiech=1;
    14     for(ll i=1;i<=100010;i++){
    15         jiech*=i;
    16         jiech%=MOD;
    17         jie[i]=jiech;
    18     }
    19 }
    20 
    21 
    22 ll jiecheng(ll aa){
    23     return  jie[aa];
    24 
    25 }
    26 
    27 ll cal(int T,int Y){
    28     if(T%2==1||Y%2==1){
    29         return 0;
    30     }
    31     ll n,m;
    32     n=T;
    33     m=(T-Y)/2;
    34     return (jiecheng(n)%MOD)/( ((jiecheng(m)%MOD)*(jiecheng(n-m)%MOD))%MOD );
    35 }
    36 
    37 int distan(int a,int b){
    38     if(a>b){
    39         return a-b;
    40     }else{
    41         return b-a;
    42     }
    43 }
    44 
    45 int main()
    46 {
    47     int n,T,w;
    48     int xi,ci;
    49     ll result=0;
    50     jiecheng1();
    51     while(~scanf("%d %d %d",&n,&T,&w)){
    52         result=0;
    53         for(int i=0;i<n;i++){
    54             scanf("%d %d",&xi,&ci);
    55             if(T==distan(xi,w)){
    56                 result+=ci;
    57             }else if(T>(int)abs(xi-w)){
    58                 result+=cal(T,distan(xi,w))*ci%MOD;
    59             }
    60         }
    61         result%=MOD;
    62         printf("%lld
    ",result);
    63 
    64 
    65     }
    66     return 0;
    67 }

    F 暂时不对

     1 #include <iostream>
     2 #include <cmath>
     3 
     4 using namespace std;
     5 
     6 int main()
     7 {
     8     double a,b,c;
     9     int n;
    10     double x,x1,x2;
    11     while(cin>>n){
    12         while(n--){
    13             cin>>a>>b>>c;
    14             if(a==0&&b==0){
    15                 if(c==0)
    16                     cout<<"NO"<<endl;
    17                 else
    18                     cout<<"YES"<<endl;
    19                 continue;
    20             }
    21             if(a==0){
    22                 x=(c*-1.0)/b;
    23                 if(x==(int)x)
    24                     cout<<"YES"<<endl;
    25                 else
    26                     cout<<"NO"<<endl;
    27                 continue;
    28             }
    29             x=(b*b-4*a*c);
    30             if(x<0){
    31                 cout<<"YES"<<endl;
    32                 continue;
    33             }
    34             x=sqrt(x);
    35             x1=(b*-1+x)/2.0*a;
    36             x2=(b*-1-x)/2.0*a;
    37             if(x1==(int)x1||x2==(int)x2)
    38                 cout<<"YES"<<endl;
    39             else
    40                 cout<<"NO"<<endl;
    41         }
    42 
    43     }
    44     return 0;
    45 }

    G

     1 package G;
     2 
     3 import java.math.BigInteger;
     4 import java.util.Scanner;
     5 
     6 public class Main {
     7     public static void main(String[] args) {
     8         int n,m;
     9         int MOD=1000000007;
    10         Scanner sc=new Scanner(System.in);
    11         while(sc.hasNext()){
    12             BigInteger iii=new BigInteger("1");
    13             BigInteger eve=new BigInteger("0");
    14             BigInteger res=new BigInteger("0");
    15             n=sc.nextInt();
    16             m=sc.nextInt();
    17             for(int i=1;i<=n;i++){
    18                 iii=iii.valueOf(i);
    19                 eve=iii.pow(m).mod(BigInteger.valueOf(MOD));
    20                 res=res.add(eve);
    21                 res=res.mod(BigInteger.valueOf(MOD));
    22             }
    23             System.out.println(res.toString());
    24         }
    25             
    26     }
    27 }

    J

     1 #include <iostream>
     2 #include <algorithm>
     3 
     4 
     5 using namespace std;
     6 
     7 struct Goods{
     8     int val,cnt;
     9 }a[1005];
    10 
    11 int cmp(Goods x,Goods y){
    12     return x.val<y.val;
    13 }
    14 
    15 int main()
    16 {
    17     int n, num, cnt;
    18     long long sum;
    19 
    20     while(cin>>n){
    21         for(int i=0;i<n;i++){
    22             cin>>a[i].val;
    23         }
    24         for(int i=0;i<n;i++){
    25             cin>>a[i].cnt;
    26         }
    27         sort(a,a+n,cmp);
    28 
    29         sum = num = 0;
    30         while(a[num].val < 0){
    31             num ++;
    32         }
    33         for(int i=num;i<n;i++){
    34             sum += a[i].val * a[i].cnt;
    35         }
    36         num--;
    37         cnt = 0;
    38 
    39         while(a[num].val + sum > 0 && num >= 0){
    40             if(cnt == 0){
    41                 cnt = a[num].cnt - 1;
    42             } else {
    43                 cnt--;
    44             }
    45             if(cnt == 0){
    46                 num--;
    47             }
    48             sum += a[num].val;
    49         }
    50 
    51         if(cnt == 0) num++;
    52         else a[num].cnt = cnt;
    53 
    54         sum = 0;
    55         for(int i=1; num < n; i++){
    56                 sum += a[num].val * i;
    57                 a[num].cnt--;
    58                 if(a[num].cnt == 0) num++;
    59         }
    60 
    61         cout<<sum<<endl;
    62     }
    63 
    64 
    65     return 0;
    66 }

    K

    #include <iostream>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    long long f[5005];
    
    struct node{
        int a,d,c;
        double f;
    };
    node sc[2005];
    
    int cmp(node x,node y){
        return x.f>y.f;
    }
    int main()
    {
        int n,T;
        while(cin>>n>>T){
            for(int i=0;i<n;i++)
                cin>>sc[i].a;
            for(int i=0;i<n;i++)
                cin>>sc[i].d;
            for(int i=0;i<n;i++){
                cin>>sc[i].c;
                sc[i].f=(sc[i].d*1.0)/sc[i].c;
            }
            sort(sc,sc+n,cmp);
            memset(f,0,sizeof(f));
            long long maxx=0;
            for(int i=0;i<n;i++){
                for(int j=T;j>=sc[i].c;j--){
                    f[j]=max(f[j],f[j-sc[i].c]+max(sc[i].a-sc[i].d*j,0));
                    maxx=max(maxx,f[j]);
                }
            }
            cout<<maxx<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/TWS-YIFEI/p/8849405.html
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