2019icpc西安邀请赛

来源：jisuanke

更新中

A.Tasks

直接贪心

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
     
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 1e6+100;
const int maxm = 6e6+100;
//const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int MAXN = maxn;
const int MAXM = maxm;
const db pi = acos(-1.0);

int n, t;
int a[maxn];
int main() {
    scanf("%d %d" ,&n ,&t);
    for(int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    sort(a,a+n);
    int tmp = 0;
    int ans = 0;
    for(int i = 0; i < n; i++){
        if(tmp+a[i]<=t){
            tmp+=a[i];ans++;
        }

    }printf("%d", ans);
    return 0;
}

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C.Angel's Journey

题意：给定一个圆的圆心(rx, ry)，半径r，A的坐标(rx, ry-r)，B的坐标(x, y)，y>ry，只能走圆上以及圆外y>ry的地方，求A到B的最短路

思路：当x<rx-r或x>rx+r的时候直接从半圆的地方走直线，否则在圆弧上走到切点然后走直线

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
     
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 1e6+100;
const int maxm = 6e6+100;
//const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int MAXN = maxn;
const int MAXM = maxm;
const db pi = acos(-1.0);

int n, t;
int a[maxn];
int main() {
    double rx,ry,r,x,y;
    int t;
    scanf("%d" ,&t);
    while(t--){
        scanf("%lf %lf %lf %lf %lf", &rx,&ry,&r,&x,&y);

        double ob = sqrt((x-rx)*(x-rx)+(y-ry)*(y-ry));
        double ans = sqrt(ob*ob-r*r)+r*(pi/2.0+asin((y-ry)/ob)-acos(r/ob));
        if(x<rx-r||x>rx+r){
            if(x<rx-r)rx-=r;
            if(x>rx+r)rx+=r;
            ans=pi/2+sqrt((x-rx)*(x-rx)+(y-ry)*(y-ry));
        }
        printf("%.4lf\n",ans);
    }
    return 0;
}

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L.Swap

题意：一个排列可以交换前n/2与后n/2，或前n^1个数奇数位置和偶数位置交换，问通过这两个操作最多产生多少个不同的排列

思路：打表发现从第五项开始是2n, n, 12, 4的规律。或者直接交上打表的模拟，由于只有两条链，而且只有一组数据，也能过

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
     
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 1e6+100;
const int maxm = 6e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n;
int a[maxn],b[maxn];
int ans;
void gao1(int a[]){
    int l = 1;
    int r = n/2+1;
    if(n&1)r++;
    for(int i = 1; i <= n/2; i++){
        swap(a[l+i-1],a[r+i-1]);
    }
    return;
}
void gao2(int a[]){
    for(int i = 1; i+1 <= n; i+=2){
        //printf("  %d %d %d\n",i,a[i],a[i+1]);
        swap(a[i],a[i+1]);
    }
    return;
}

int sv(int n){
    ::n=n;
    ans=1;
    int sta=1;
    for(int i = 1; i <= n; i++)a[i]=b[i]=i;
    /*if(n==1)return 1;
    if(n==2)return 2;
    if(n==3)return 6;*/
    gao1(a);gao2(b);
    //for(int i = 1; i <= n; i++)printf("%d ",a[i]);printf("\n");
    //for(int i = 1; i <= n; i++)printf("%d ",b[i]);printf("\n");
    while(1){
        //for(int i = 1; i <= n; i++)printf("%d ",a[i]);printf("\n");
        //for(int i = 1; i <= n; i++)printf("%d ",b[i]);printf("\n");
        int ys = 0;
        sta^=1;
        for(int i = 1; i <= n; i++){
            if(a[i]!=b[i])ys=1;
        }
        if(!ys){
            ans++;break;
        }
        else{
            ans+=2;
            if(sta) {gao1(a);gao2(b);}
            else {gao1(b);gao2(a);}
        }
    }
    return ans;
}
int main() {
    //scanf("%d" ,&n);
    //sv(3);
    for(int i = 1; i <= 100; i++){
        printf("%d %d\n",i,sv(i));
    }
    scanf("%d", &n);
    //for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
        printf("%d",sv(n));
    return 0;
}

View Code

2019icpc西安邀请赛

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