AtCoder Beginner Contest 113

A - Discount Fare

$z=x+y/2$

include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 
int x,y,z;
int main()
{
	while(scanf("%d%d",&x,&y)!=EOF)
	{
		z = y / 2 + x;
		cout<<z<<endl;
	}
}

B - Palace

比较差值的绝对值的大小

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 
int n,T,a;
struct node 
 {
 	ll tt;
 	int id;
 }t[2110];
 
 bool cmp(node a,node b)
 {
 	return a.tt <= b.tt;
 }
int x;
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		cin>>T>>a;
		a *= 1000;
		rep(i,0,n)
		{
			cin>>x;
			t[i].tt = 1000 * T - x * 6;
			t[i].tt = abs(a - t[i].tt);  
			t[i].id = i+1;
		}
		sort(t,t+n,cmp);
		cout<<t[0].id<<endl;
	}
}

C - ID

先整个进行时间的排序，再遍历一遍

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 1e5+1000;
 
 
int n,m,a;
struct node 
 {
 	ll tt;
 	int pp;
 	int id;
 } t[maxn];
 
 bool cmp(node a,node b)
 {
 	return a.tt <= b.tt;
 }
ll p,y;
 
int s[maxn];
int pre[maxn];
int  en[maxn];
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(s,0,sizeof s);
		rep(i,0,m)
		{
			cin>>p>>y;
			t[i].tt = y; 
			t[i].pp = p;
			t[i].id = i;
		}
		sort(t,t+m,cmp);
		rep(i,0,m)
		{
			pre[t[i].id] = t[i].pp;
			en[t[i].id] = ++s[t[i].pp]; 
		}
		rep(i,0,m)
		{
			printf("%06d%06d\n",pre[i],en[i]);
		}
 
	}
}

D - Number of Amidakuji

注意题目中K代表的意义，K是指从左边开始沿着边走最后到达的第几根，注意可以往回走（可以看第四个样例第一个介绍）

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 
int h,w,k;
ll dp[1000][110];
int s[] = {1,1,2,3,5,8,13,21};
 
int main(int argc, char const *argv[])
{
	cin>>h>>w>>k;
	if(w == 1)
		{
			printf("1\n");
			return 0;
		}
	dp[0][0] = 1;
	rep(i,0,h)
		rep(j,0,w)
		{
			dp[i][j] %= mod;
			if(j == 0)
			{
				dp[i+1][j+1] += dp[i][j] * s[max(0,w - j - 2)];
				dp[i+1][j] += dp[i][j] * s[w - j - 1];
			}
			else if(j != 0)
			{
				dp[i+1][j+1] += dp[i][j] * s[j] * s[max(0,w - j - 2)];
				dp[i+1][j] += dp[i][j] * s[j] * s[w - j - 1];
				dp[i+1][j-1] += dp[i][j] * s[max(0,j-1)] * s[w-j-1];
			}
		}		
	cout<<dp[h][k-1]%mod<<endl;	
	return 0;
}