codeforces - 1245 (div2)

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A - Good ol‘ Numbers Coloring

直接判断两个数是否互质

技术图片
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 1e6 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, a, b;
 
int gcd(int a,int b){ return b==0?a:gcd(b,a%b); }
 
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    cin >> t;
    while (t--) {
        cin >> a >> b;
        if (gcd(a, b) == 1)    puts("Finite");
        else    puts("Infinite");
    }
}
View Code

 

B - Restricted RPS 

贪心

技术图片
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 200, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int t, a, b, c, flag, tmp, m, n;
char s1[N], s2[N];
 
int gcd(int a,int b){ return b==0?a:gcd(b,a%b); }
 
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    cin >> t;
    while (t--) {
        tmp = 0;
        mm(s2, 0);
        cin >> n >> a >> b >> c;
        scanf("%s", s1);
        for (int i = 0; i < n; ++i) {
            if (s1[i] == R) {
                if (b > 0) {
                    tmp++;
                    b--;
                    s2[i] = P;
                } else {
                    s2[i] = X;
                }
            } else if (s1[i] == P) {
                if (c > 0) {
                    c--;
                    tmp++;
                    s2[i] = S;
                } else {
                    s2[i] = X;
                }
            } else if (s1[i] == S) {
                if (a > 0) {
                    tmp++;
                    a--;
                    s2[i] = R;
                } else {
                    s2[i] = X;
                }
            }
        }
        if (n & 1)    m = n / 2 + 1;
        else    m = n / 2;
        
//        cout << tmp << "  " << m << endl;
        
        if (tmp < m) {
            puts("NO");
            continue;
        } else {
            for (int i = 0; i < n; ++i) {
                if (s2[i] == X) {
                    if (a > 0) {
                        s2[i] = R;
                        a--;
                    } else if (b > 0) {
                        s2[i] = P;
                        b--;
                    } else if (c > 0) {
                        c--;
                        s2[i] = S;
                    }
                }
            }
            printf("YES
");
            printf("%s
", s2);
        }
        
    }
}
/*
7
2
1 1 0
RR
2
2 0 0
SS
2
2 0 0
RR
3
1 2 0
RRR
3
2 1 0
RRR
1
1 0 0
P
1
1 0 0
S
*/
View Code

 

C. Constanze‘s Machine

如果字符中出现m或w就代表是由好的打印机发过来的,直接输出0。否则就dp推一下状态转移方程,后来发现如果当前字符为n或u并且前一个字符也为n或u就是dp[n] = dp[n - 1] + dp[n - 2],否则dp[n] = dp[n - 1]

技术图片
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 1e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
char s[N];
int dp[N];
int ans, flag;
 
int pow_2(int x) {
    int ans = 2;
    for (int i = 1; i <= x; ++i)
        ans *= 2;
    return ans;
}
 
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    scanf("%s", s);
    flag = 0;
    int len = strlen(s);
    dp[0] = 1;
    for (int i = 0; i < len; ++i) {
        if (s[i] == m || s[i] == w) {
            flag = 1;
            break;
        }
        if (i == 0)    continue;
        if (i == 1) {
            if (s[i] == s[i - 1])    dp[i] = dp[i - 1] + 1;
            else dp[i] = dp[i - 1];
            continue;
        }
        
        if (s[i] == u && s[i - 1] == u)        dp[i] = (dp[i - 1] + dp[i - 2]) % mod;
        else if (s[i] == n && s[i - 1] == n)    dp[i] = (dp[i - 1] + dp[i - 2]) % mod;
        else dp[i] = dp[i - 1];
        
//        if (s[i] == ‘n‘) {
//            if (i == 1) {
//                if (s[i - 1] == ‘n‘) dp[i] = dp[i - 1] * 2;
//                else dp[i] = dp[i - 1];
//                continue;
//            }
//            if (s[i - 1] == ‘n‘ && s[i - 2] == ‘n‘)    dp[i] = dp[i - 1] + dp[i - 2];
//            else if (s[i - 1] == ‘n‘)    dp[i] = (dp[i - 1] + 1) * dp[i - 1];
//            else    dp[i] = dp[i - 1];
//        } else if (s[i] == ‘u‘) {
//            if (i == 1) {
//                if (s[i - 1] == ‘u‘) dp[i] = dp[i - 1] + 1;
//                else dp[i] = dp[i - 1];
//                continue;
//            }
//            if (s[i - 1] == ‘n‘ && s[i - 2] == ‘u‘)    dp[i] = dp[i - 1] + dp[i - 2];
//            else if (s[i - 1] == ‘u‘)    dp[i] = dp[i - 1] + 1;
//            else    dp[i] = dp[i - 1];
//        } else {
//            dp[i] = dp[i - 1];
//        }
//        
//        cout << i << " " << dp[i] << endl;
    }
    if (flag == 1) {
        puts("0");
    } else {
        cout << dp[len - 1] << endl;
    }
}
View Code

 

D - Shichikuji and Power Grid

题意:要使每个村庄都通上电,通电的方法是在这个村庄建发电站(花费ci)或者把这个村庄和一个有发电站的村庄连接起来(花费ki + kj的和乘上两个村庄的曼哈顿距离)

这题显然是最小生成树(感觉迪杰斯特拉也行),但是n比较小,可以直接暴力贪心

解法:首先假设每个点都自建,那么每个点的代价就是自建代价。然后按照代价排序,用代价最小的点去更新后面那些点,如果能更新用电代价,就把那些点连接到当前点。然后进入下一轮循环,排除上一次代价最小的点,把剩下的点再次按照代价排序,然后用这些点中代价最小的去更新其他的,以此类推。

技术图片
By MengWH, contest: Codeforces Round #597 (Div. 2), problem: (D) Shichikuji and Power Grid, Accepted, #
#include<bits/stdc++.h>
using namespace std ;
int n;
struct City {
    int id;
    long long x,y;  //坐标
    long long cc,kk;  //自建的花费,连线的花费
    bool self;//是否建站
    int fa;//连线的站
    bool operator < (const City & a)const {
        return cc<a.cc;
    }
} c[2005];
int main() {
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        c[i].id=i;//发电站编号
        c[i].self=1;  //首先都默认是自建的
        scanf("%lld%lld",&c[i].x,&c[i].y);  //输入坐标
    }
    for(int i=1; i<=n; i++) scanf("%lld",&c[i].cc);  //初始都为自建
    for(int i=1; i<=n; i++) scanf("%lld",&c[i].kk);//连线
    long long ans=0,selfnum=0;
    for(int i=1; i<=n; i++) {
        sort(c+i,c+1+n);//大概就是要随时排序,每次找到最小的,每次排序要排除前一次的,防止多加费用
        ans+=c[i].cc;  //费用
        if(c[i].self) selfnum++;  //判断是否自建
        for(int j=i+1; j<=n; j++) {
            long long cost=(c[i].kk+c[j].kk)*(abs(c[i].x-c[j].x)+abs(c[i].y-c[j].y));
            if(cost<c[j].cc) {
                c[j].cc=cost;
                c[j].self=0;//放弃自建,说要已经和别的站建立了联系
                c[j].fa=c[i].id;
            }
        }
    }
    printf("%lld
%lld
",ans,selfnum);
    for(int i=1; i<=n; i++)
        if(c[i].self) printf("%d ",c[i].id);
    printf("
%lld
",n-selfnum);
    for(int i=1; i<=n; i++)
        if(!c[i].self) printf("%d %d
",c[i].id,c[i].fa);
    return 0;
}
View Code

 

E、F

......

 

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