2020-2021年度第二届全国大学生算法设计与编程挑战赛（秋季赛）-正式赛-详细题解

以你之姓@ 2022-12-22 00:47 190阅读 0赞

### 文章目录 ###

*  A小x的奇遇-adventure
 *  B数位dp-dp
 *  F大数据分析-data

# A小x的奇遇-adventure #

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#include <cstdio>
    
    const int N = 1e6 + 5;
    int tot, p[N];
    bool flg[N];
    
    void sieve(int n){
        
    	for(int i=2; i<=n; i++){
        
    		if(!flg[i]){
        
    			p[++tot] = i;
    		}
    		for(int j=1; j<=tot&&i*p[j]<=n; ++j){
        
    			flg[i*p[j]] = 1;
    			if(i%p[j]==0){
        
    				break;
    			}
    		}
    	}
    } 
    
    long long phi(long long x){
        
    	long long ans = x;
    	for(int i=1; i<=tot&& 1ll*p[i]*p[i]; ++i){
        
    		if(x%p[i]){
        
    			continue;
    		}
    		ans = ans/p[i]*(p[i]-1);
    		while(x%p[i]==0){
        
    			x /= p[i];
    		}
    	}
    	if(x>1){
        
    		ans = ans / (x * (x - 1));
    	}
    	return ans;
    }
    
    int main() {
        
    	sieve(N - 5);
    	long long n, k;
    	scanf("%llld%lld", &n, &k);
    	k = (k + 1) / 2;
    	for(long long i=1; i<=k&&n>1; ++i){
        
    		n = phi(n);
    	}
    	printf("%lld\n", n%1000000007);
    	
    	return 0;
    }

# B数位dp-dp #

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#include <iostream>
    #include <cmath>
    using namespace std;
    int work(int x){
        
    	int L = 0, R = -1, lst = -1; //L最高位 R最低位 lst上一位数 
    	while(x){
        
    		int tmp = x % 10;
    		L = tmp;  
    		if(R==-1){
        
    			R = tmp;
    		}
    		/*相邻来两位>7*/ 
    		if(lst!=-1){
        
    			if(abs(tmp-lst)>7){
        
    				return 0;
    			}
    		}
    		lst = tmp;
    		x /= 10;
    	}
    	/*最高位和最低位差值>2*/
    	return abs(R-L)>2;
    }
    
    int main(){
        
    	int cnt = 0;
    	for(int i=13930; i<=457439; i++){
        
    		cnt += work(i);
    	}
    	cout<<cnt;
    	return 0;
    }

# F大数据分析-data #

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#include <iostream>
    using namespace std;
    int main(){
        
    	//ans累计，now当前的那个数 
    	long long ans = 1, now = 1, MOD = 1000000007ll;
    	for(int i=2; i<=30; i++){
        
    		now = now * (now + 5);
    		now %= MOD;
    		ans += now;
    		ans %= MOD;
    	}
    	cout << ans;
    	return 0;
    }

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