标签：hdu   读入外挂   set   

Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 502    Accepted Submission(s): 141

2
1 1
1
2 1
-1 0

Case #1: Yes.
Case #2: No.

Hint
If input is huge, fast IO method is recommended.
 

#include <cstdio>
#include <cstring>
#include <set>
#define ll long long
using namespace std;
int const MAX = 1e6 + 5;
ll a[MAX], sum[MAX];
set <ll> s;

ll Scan()  
{  
    ll res = 0, ch, flag = 0;  
    if((ch = getchar()) == '-')           
        flag = 1;  
    else if(ch >= '0' && ch <= '9')         
        res = ch - '0';  
    while((ch = getchar()) >= '0' && ch <= '9' )  
        res = res * 10 + ch - '0';  
    return flag ? -res : res;  
}  

int main()
{
    int T, n, k;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        bool flag = false;
        s.clear();
        memset(sum, 0, sizeof(sum));
        scanf("%d %d", &n, &k);
        getchar();
        for(int i = 1; i <= n; i++)
            a[i] = Scan();
        for(int i = 1; i <= n; i++)
            sum[i] = sum[i - 1] + (i % 2 ? a[i] : -a[i]);
        for(int i = n; i > 0; i--)
        {
            s.insert(sum[i]);
            if(i % 2)
            {
                if(s.find(sum[i - 1] + k) != s.end())
                {
                    flag = true;
                    break;
                }
            }
            else
            {
                if(s.find(sum[i - 1] - k) != s.end())
                {
                    flag = true;
                    break;
                }
            }
        }
        printf("Case #%d: %s.\n", ca, flag ? "Yes" : "No");
    }
}

HDU 5183 Negative and Positive (NP) (set ＋ 读入外挂 乱搞)

标签：hdu   读入外挂   set   

原文地址：http://blog.csdn.net/tc_to_top/article/details/44127119