Negative and Positive (NP) （hdu 5183 set+输入外挂）

Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2488    Accepted Submission(s): 618

Problem Description

When given an array (a0,a1,a2,⋯an−1) and an integer K, you are expected to judge whether there is a pair (i,j)(0≤i≤j<n) which makes that NP−sum(i,j) equals to K true. Here NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj

 

Input

Multi test cases. In the first line of the input file there is an integer T indicates the number of test cases.
In the next 2∗T lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers n and K which are mentioned above.
The second line contain (a0,a1,a2,⋯an−1)separated by exact one space.
[Technical Specification]
All input items are integers.
0<T≤25,1≤n≤1000000,−1000000000≤ai≤1000000000,−1000000000≤K≤1000000000

 

Output

For each case，the output should occupies exactly one line. The output format is Case #id: ans, here id is the data number starting from 1; ans is “Yes.” or “No.” (without quote) according to whether you can find (i,j) which makes PN−sum(i,j) equals to K.
See the sample for more details.

 

Sample Input

21 112 1-1 0

 

Sample Output

Case #1: Yes.Case #2: No.
Hint
If input is huge, fast IO method is recommended.
 

 

Source

BestCoder Round #32

 

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题意：问有没有数对(i,j)(0<=i<=j<n),使得a[i]-a[i+1]+...+(-1)^(j-i)a[j]为K.

思路：先计算出前缀和sum，然后枚举起点，从后往前枚举起点i，若i为奇数，则看set里面有没有sum[i-1]+k；若i为偶数，则看set里面有没有sum[i-1]-k。要用到输入外挂，不知道为什么，我的代码用G++交有时能过，有时却TLE，难道还要看OJ的心情吗=。=

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005000#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FRL(i,a,b)  for(i = a; i < b; i++)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;ll sum[maxn];ll a[maxn];ll n,k;set<ll>s;inline bool scan_d(ll &ret) //输入外挂{    char c;    ll sgn;    if (c=getchar(),c==EOF) return false;    while (c!='-'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while (c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    ret*=sgn;    return true;}int main(){    ll x;    ll i,j,t,cas=1;    scan_d(t);    while (t--)    {        scan_d(n);        scan_d(k);        FRL(i,1,n+1)            scan_d(a[i]);        sum[0]=0;        sum[1]=a[1];        FRL(i,2,n+1)    //计算前缀和        {            if (i%2)                sum[i]=sum[i-1]+a[i];            else                sum[i]=sum[i-1]-a[i];        }        s.clear();        bool flag=false;        for (i=n;i>0;i--)        {            s.insert(sum[i]);   //现插入集合            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;                }            }        }        if (flag)            pf("Case #%lld: Yes.\n",cas++);        else            pf("Case #%lld: No.\n",cas++);    }    return 0;}