2019.7.9

Posted 2022-04-01 fanshhh

2019.7.9

cf #572 div2 A

给你一个只有0和1的字符串，如果0和1的数目不相等，那么就是good，求用最少次数拆分字符串，使得他所有子串都是good。

思路：如果字符串是奇数位的，显然该字符串是good，不用拆分。如果是偶数位，先判断字符串是否是good，如果不是good，那就把第一个字符拆出来。

#include <cstdio>

int main() 
    int n;
    char a[110];
    while(~scanf("%d %s", &n, a)) 
        int cnt = 0;
        if(n % 2) printf("1\\n%s\\n", a);
        else 
            for(int i = 0; i < n; i++) 
                if(a[i] == '0') cnt++;
            
            if(cnt != n / 2) printf("1\\n%s\\n", a);
            else 
                printf("2\\n%c ", a[0]);
                for(int i = 1; i < n; i++) printf("%c", a[i]);
                printf("\\n");
            
        
    
    return 0;

cf #572 div2 B

#include <cstdio>
#include <algorithm>
using namespace std;
int main() 
    int n;
    while(~scanf("%d", &n)) 
        int a[100010];
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        sort(a,a + n);
        if(a[n-1] >= a[n - 2] + a[n - 3]) printf("NO\\n");
        else 
            printf("YES\\n");
            for(int i = 0; i < n; i += 2) printf("%d ", a[i]);
            for(int i = n - 1 - (n%2); i >= 0; i -= 2) printf("%d ", a[i]);
            printf("\\n");
        
    
    return 0;

cf #572 div2 C

https://www.cnblogs.com/wyboooo/p/11153973.html

#include <cstdio>
#include <algorithm>
using namespace std;
int main() 
    int n;
    while(~scanf("%d", &n)) 
        int a[100010];
        a[0] = 0;
        for(int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
            a[i] += a[i-1];
        
        int q;
        scanf("%d", &q);
        while(q--) 
            int l, r;
            scanf("%d %d", &l, &r);
            printf("%d\\n", (a[r] - a[l-1]) / 10);
        
    
    return 0;

cf #572 div2 D1

判断是否存在度为二的节点，如果有就NO，否则YES。

题意：能否通过改变两个叶子之间所有路径的值，使得每条路径可以任意改变其值。如果有度为二的节点，那么就有两条边的值是一定相等的。（注：题目中只有4567是叶子）

#include <cstdio>
#include <cstring>
int main() 
    int n;
    while(~scanf("%d", &n)) 
        int a[100010], l, r, flag = 0;
        memset(a, 0, sizeof(a));
        for(int i = 0; i < n - 1; i++) 
            scanf("%d %d", &l, &r);
            a[l]++, a[r]++;
        
        for(int i = 0; i < 100010; i++) 
            if(a[i] == 2) 
                flag = 1;
                break;
            
        
        if(flag) printf("NO\\n");
        else printf("YES\\n");
    
    return 0;

cf #572 div2 E

https://www.cnblogs.com/wyboooo/p/11155925.html

#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long LL;
using namespace std;

int a[300010];
LL c[300010];

int main() 
    c[0] = 0;
    for(int i = 1; i < 300010; i++) c[i] = c[i-1] + i;
    LL n, p, k;
    while(~scanf("%lld %lld %lld", &n, &p, &k)) 
        memset(a, 0, sizeof(a));
        LL ans = 0, b;
        for(int i = 0; i < n; i++) 
            scanf("%lld", &b);
            b = (b % p * b % p * b % p * b % p) + p - ((b % p) * (k % p) ) % p;
            b = b % p;
            a[i] = b;
        
        sort(a, a+n);
        a[n] = p;
        int cnt = 1;
        for(int i = 0; i < n; i++) 
            if(a[i] == a[i+1]) 
                cnt++;
            
            else 
                ans += c[cnt-1];
                cnt = 1;
            
        
        printf("%lld\\n", ans);
    
    return 0;

cf #571 div2 A

水题

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    if(min(b,c)>=a) printf("Yes\\n");
    else printf("No\\n");
    return 0;

cf #571 div2 D

注意9.000只能是9，而不能是10

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
    int n;
    while(~scanf("%d", &n)) 
        double a[100010];
        long long ans = 0;
        for(int i = 0; i < n; i++) 
            scanf("%lf", &a[i]);
            ans += (int)a[i];
        
        //printf("ans=%d\\n",ans);
        if(ans >= 0) 
            for(int i = 0; i < n; i++) 
                if(ans > 0 && a[i] <= 0 &&( (long long)(a[i] * 100000) % 100000 )!= 0) 
                    printf("%d\\n",(int) a[i] - 1);
                    ans--;
                
                else printf("%d\\n",(int)a[i]);
            
        
        else 
            for(int i = 0; i < n; i++) 
                if(ans < 0 && a[i] >= 0 &&( (long long)(a[i] * 100000) % 100000 )!= 0) 
                    printf("%d\\n", (int)a[i] + 1);
                    ans++;
                
                else printf("%d\\n",(int)a[i]);
            
        
    
    return 0;