吉首大学第八届“新星杯”大学生程序设计大赛（暨新生网络同步赛）

A-组合数

#include<stdio.h>
#include<string.h>
#define ll long long          
const int N=100;
const int mod=1e9+7;
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,q;
ll a[505]={1,1},b[500];
ll dp[450][450],cnt;
ll c[N][N];
void init()
{
	ll i,j;
	for(i=2;i<=432;i++)
	{
		if(a[i])
			continue;
		b[++cnt]=i;
		for(j=i*i;j<=432;j+=i)
			a[j]=1;
	}
	for(i=2;i<=432;i++)
	{
		for(j=1;b[j]<=i && j<=cnt;j++)
		{
			ll tmp=i,res=0;
			while(tmp)
				tmp/=b[j],res+=tmp;
			dp[i][b[j]]=res;
		}
	}
}

int main(){
	c[0][0]=1;
	for(int i=1;i<=50;i++){
		c[i][0]=c[i][i]=1;
		for(int j=1;j<i;j++){
			c[i][j]=c[i-1][j]+c[i-1][j-1];
		}
	}
	ll n,m;
	init();
	while(~scanf("%lld%lld",&n,&m)){
		ll ans=1;
		for(int i=1;b[i]<=n&&i<=cnt;i++){
			ll tmp=dp[n][b[i]]-dp[m][b[i]]-dp[n-m][b[i]];
			ans*=(tmp+1);
		}
		printf("%lld %lld\n",c[n][m],ans);
	}
	return 0;
}

B-傻鸡抓大闸蟹

#include<bits/stdc++.h>
using namespace std;
int vis[5010];
struct node{
	long long x,y;
	int id;
}a[5100];
int cmp(node x1,node y1){
	return x1.x<y1.x;
}
int main(){
	int n;
	memset(vis,0,sizeof(vis));
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i].x,&a[i].y);
		a[i].id=i;
	}
	sort(a+1,a+1+n,cmp);
	for(int i=1;i<=n/2;i++){
		vis[i]=1;
		vis[n-i+1]=1;
		if(a[i].id>a[n-i+1].id)
		swap(a[i].id,a[n-i+1].id);
		printf("%d\n",a[i].id);
		printf("%d\n",a[n-i+1].id);
	}
	for(int i=1;i<=n;i++){
		if(vis[i]==0)
		printf("%d\n",a[i].id);
	}
}

问题 E: Jack的A+B

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int main(){
	long long a,b;
	stack<int>s;
	while(~scanf("%lld%lld",&a,&b)){
		if(a+b==0)
		{
			printf("0\n");
			continue;
		}
		long long ans=a+b;
		int t,ct=0;
		while(ans>0){
			t=ans%10;
			s.push(t);
			ct++;
			if(ct==3&&ans/10>0){
				ct=0;
				s.push(11);
			}
			ans/=10;
		}
		while(!s.empty()){
			if(s.top()==11){
				printf(",");
				s.pop();
			}
			else{
			printf("%d",s.top());
			s.pop();
		} 	
	}
		printf("\n");
}
}

问题 G: 圆桌上的晚餐

#include<stdio.h>
#include<string.h>
int main(){
	int n,m;
	int a[1005];
	while(~scanf("%d%d",&n,&m)){
		int f=0;
		memset(a,0,sizeof(a));
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		if(a[m-1]==0){
			printf("%d\n",m);
			continue;
		}
		else{
			int t=m-1;
			while(t++<n-1){
				if(a[t]==0){
				f=1;
				printf("%d\n",t+1);
				break;
				}
			
			}
			if(f==0){
				for(int i=0;i<m-1;i++){
					if(a[i]==0){
						printf("%d\n",i+1);
						break;
					}
				}
			}
		}
	}
}

问题 H: 小k的简单问题

#include<bits/stdc++.h>
using namespace std;
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int x,y,z;
		int n;
		scanf("%d",&n);
		int sum1=0,sum2=0,sum3=0;
			for(int i=0;i<n;i++){
				scanf("%d%d%d",&x,&y,&z);	
				sum1+=x*z;
				sum2+=y*z;
				sum3+=z;
		}
		printf("%.2lf %.2lf\n",(double)sum1/sum3,(double)sum2/sum3);
	}
}

问题 I: 夫子云游

#include<cstdio>
int n,m,ans;
void dfs(int a,int b,int c,int d){
	if(b>=n+m){
		if(a==0)
		ans++;
		return;
	}
	if(a<=0)
	return;
	if(c<n)
	dfs(a-1,b+1,c+1,d);
	if(d<m)
	dfs(a*2,b+1,c,d+1);
}
int main(){
	scanf("%d%d",&m,&n);
		ans=0;
		dfs(2,0,0,0);
	printf("%d\n",ans);
	
} 

问题 J: 老肖数等式

#include<stdio.h>
int main(){
	int n;
	while(~scanf("%d",&n)){
		for(int i=0;i<=n/2;i++){
			printf("%d. %d+%d=\n",i+1,i,n-i);
		}
	}
}

问题 K: WaWa的难题

#include<bits/stdc++.h>
using namespace std;
int N;
int vis[100];
int ans;
int main(){
	long long a[105];
	a[0]=0;
	a[1]=1;
	a[2]=1;
	a[3]=2;
	a[4]=4;
	for(int i=5;i<=100;i++)
	a[i]=a[i-1]+a[i-3]+1;
	while(~scanf("%d",&N)){
		printf("%lld\n",a[N]);
	}
}

问题 M: 最亲密的x个人

#include<bits/stdc++.h>
using namespace std;
struct node{
	long long x2=0;
	int id=0;
	bool operator <(const node &S)const{
        if(x2==S.x2)
            return id<S.id;
        return x2<S.x2;
    }

}a[100005];
int main(){
	int n,x,l;
	long long m,t;
	int cnt;
	while(~scanf("%d%lld%d",&n,&m,&x)){
		cnt=0,l=1;
		for(int i=1;i<=n;i++){
			scanf("%lld",&t);
			a[i].id=i;
			a[i].x2=abs(t-m);
		}
		sort(a+1,a+1+n);
		while(cnt<x){
			if(a[l].x2!=0){
				long long q=a[l].x2;
				while(a[l].x2==q&&cnt<=x){
					printf("%d\n",a[l].id);
					l++;
					cnt++;
				}
			}
			else l++; 
		}
	}
	return 0;
}