足夠 scan property ++ inter digits pos 通過 題意

C. Vasya and Golden Ticket time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Recently Vasya found a golden ticket — a sequence which consists of $\mathrm{nn\; digits\; a1a2\dots ana1a2\dots an.\; Vasya\; considers\; a\; ticket\; to\; be\; lucky\; if\; it\; can\; be\; divided\; into\; two\; or\; more\; non-intersecting\; segments\; with\; equal\; sums.\; For\; example,\; ticket}$

350178350178 is lucky since it can be divided into three segments $\mathrm{350350,\; 1717\; and\; 88:\; 3+5+0=1+7=83+5+0=1+7=8.\; Note\; that\; each\; digit\; of\; sequence\; should\; belong\; to\; exactly\; one\; segment.}$

Help Vasya! Tell him if the golden ticket he found is lucky or not.

Input

The first line contains one integer $\mathrm{nn\; (2\le n\le 1002\le n\le 100)\; —\; the\; number\; of\; digits\; in\; the\; ticket.}$

The second line contains $\mathrm{nn\; digits\; a1a2\dots ana1a2\dots an\; (0\le ai\le 90\le ai\le 9)\; —\; the\; golden\; ticket.\; Digits\; are\; printed\; without\; spaces.}$

Output

If the golden ticket is lucky then print "YES", otherwise print "NO" (both case insensitive).

Examples input Copy

5
73452

output Copy

YES

input Copy

4
1248

output Copy

NO

Note

In the first example the ticket can be divided into $\mathrm{77,\; 3434\; and\; 5252:\; 7=3+4=5+27=3+4=5+2.}$

In the second example it is impossible to divide ticket into segments with equal sum.

【題意】：

　　題意的大概意思是是不是存在兩組以上的序列使得他們相等，比如 773434 可以等價於 7 = 7 = 3 + 4 = 3 + 4

【思路】：

　　我的思路是先求一次前綴和，然後我們要判斷的是存不存在大於2組的序列，那麽前綴和sum[n] = x * n；x是滿足的序列

那麽我們只要假設x，然後翻倍向上求就可以了，翻倍查詢是不是存在，x * 1， x * 2， x * 3 -----x * n這樣的前綴和，因為

這個的總和很少，所以我們可以開一個足夠大的標記數組來標記sum[i],,,,,,sum[n]的數值，這樣的話，查詢的復雜度就是O（1）了

我們只要枚舉sum[i] ----sum[m],如果sum[m] > sum[n] / 2的話就不用判斷了，因為肯定不存在答案，大概復雜度是O（n * log n sum[i]）

因為數據量不大，所以可以通過

全都是0的情況需要特判

附上代碼：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int flag[2000];
int math[2000];
int sum[2000];
char str[1005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(flag, 0, sizeof(flag));
        memset(math, 0, sizeof(math));
        memset(sum, 0, sizeof(sum));
        memset(str, 0, sizeof(str));
        scanf("%s",str);
        for(int i = 0; i < n; i ++)
        {
            math[i] = str[i] - ‘0‘;
        }
        sum[0] = math[0];
        flag[sum[0]] = 1;
        for(int i = 1; i < n; i ++)
        {
            sum[i] = sum[i - 1] + math[i];
            flag[sum[i]] ++;
        }
        if(flag[0] >= 2 && sum[n - 1] == 0)
        {
            printf("YES\n");
            continue;
        }
        else
        {
            int nums = sum[n - 1] / 2 + 1;
            int cs = 0;
            for(int i = 0; i < n; i ++)
            {
                if(sum[i] == 0)
                    continue;
                if(cs)
                    break;
                if(sum[i] > nums)
                    break;
                else
                {
                    int summ = 0;
                    int fff = 1;
                    int ff = 0;
                    while(summ < sum[n - 1])
                    {
                        //printf("...\n");
                        summ = summ + sum[i];
                        ff ++;
                        if(flag[summ] >= 1 && summ == sum[n - 1] && ff >= 2)
                        {
                            fff = 0;
                            break;
                        }
                        if(!flag[summ])
                        {
                            fff = 1;
                            break;
                        }
                    }
                    //printf("%d %d\n",ff,fff);
                    if(!fff)
                    {
                        printf("YES\n");
                        cs = 1;
                        break;
                    }
                }
            }
            if(!cs)
                printf("NO\n");
        }
    }
    return 0;
}

Codeforces Round #512 (Div. 2, based on Technocup 2019 Elimination Round 1) C. Vasya and Golden Ticket