Codeforces Round #359 (Div. 2)

难得AB出的还算比较快，被C的0~n-1调了1h，一首凉凉送给自己。

 

作者太菜，所以只有前三题的题解，抱歉。 

 

A. Free Ice Cre

Sol：

直接按题意模拟即可

#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#define rg register
#define il inline
#define vd void
#define ll long long
#define For(i,x,y) for (rg int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (rg int i=(x);i>=(y);i--)
#define cross(i,k) for (rg int i=first[k];i;i=last[i])
using namespace std;
il ll max(ll x,ll y){return x>y?x:y;}
il ll min(ll x,ll y){return x<y?x:y;}
il ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int n,x,ans;
ll m;
char c[20];
int main(){
    n=read(),m=read();
    For(i,1,n){
        scanf("%s",c),x=read();
        if (c[0]=='+') m+=1ll*x;
        else if (m<x) ans++;
        else m-=x;
    }
    printf("%lld %d",m,ans);
}

 

B.Little Robber Girl's Zoo

Sol：

n只有100，根据冒泡排序我们可以知道，最多交换n^2次，而题目里的上限是2w。

所以直接模拟。

#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#define rg register
#define il inline
#define vd void
#define ll long long
#define N 101
#define For(i,x,y) for (rg int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (rg int i=(x);i>=(y);i--)
#define cross(i,k) for (rg int i=first[k];i;i=last[i])
using namespace std;
il ll max(ll x,ll y){return x>y?x:y;}
il ll min(ll x,ll y){return x<y?x:y;}
il ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int n,x,a[N],b[N];
int main(){
    n=read();
    For(i,1,n) a[i]=b[i]=read();
    sort(b+1,b+1+n);
    Dow(i,n,1){
        if (a[i]==b[i]) continue;
        Dow(j,i-1,1)
            if (a[j]==b[i]){
                x=j;
                break;
            }
        For(j,x,i-1){
            printf("%d %d\n",j,j+1);
            swap(a[j],a[j+1]);
        }
    }
}

 

 

C.Robbers' watch

Sol：

因为不能有重复的数出现，而且一共就7个数字，所以n-1转成七进制的位数+m-1转成七进制的位数大于7就直接输出0，否则dfs。

#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#define rg register
#define il inline
#define vd void
#define ll long long
#define For(i,x,y) for (rg int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (rg int i=(x);i>=(y);i--)
#define cross(i,k) for (rg int i=first[k];i;i=last[i])
using namespace std;
il ll max(ll x,ll y){return x>y?x:y;}
il ll min(ll x,ll y){return x<y?x:y;}
il ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,ans,a[20],b[20];

bool vis[10];
il vd Dfs(int k,bool flag){
    if (k>b[0]){ans++;return;}
    int l=flag?b[b[0]-k+1]:6;
    For(i,0,l)
        if (!vis[i]){
            vis[i]=1;
            Dfs(k+1,flag&&i==b[b[0]-k+1]?1:0);
            vis[i]=0;
        }
}
il vd dfs(int k,bool flag){
    if (k>a[0]){Dfs(1,1);return;}
    int l=flag?a[a[0]-k+1]:6;
    For(i,0,l)
        if (!vis[i]){
            vis[i]=1;
            dfs(k+1,flag&&i==a[a[0]-k+1]?1:0);
            vis[i]=0;
        }
}

int main(){
    n=read()-1,m=read()-1;
    if (!n) a[0]=1; 
    while (n) a[++a[0]]=n%7,n/=7;
    if (!m) b[0]=1;
    while (m) b[++b[0]]=m%7,m/=7;
    if (a[0]+b[0]>7) printf("0");
    else dfs(1,1),printf("%d",ans);
}