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but he remembers that among thescores announced

时间:2018-08-09 04:14来源:雪儿 作者:伊沙 点击:
coutans; return 0; break; ans--; if(!binary_search(a+1,a+n+1,b[p]-b[1]+a[j]))//自带的二分函数 //he增量delta =b[p] - b[1] + a[j] //b[p]-x在a[]中能找得到 二分解决 //sex x rated韩国妈妈如果x可行,返回去重后的最后元素的地址) for(int p=2;p=k;p

   cout<<ans;

return 0;

break;

ans--;

if(!binary_search(a+1,a+n+1,b[p]-b[1]+a[j]))//自带的二分函数

//he增量delta =b[p] - b[1] + a[j]

//b[p]-x在a[]中能找得到 二分解决

//sex x rated韩国妈妈如果x可行,返回去重后的最后元素的地址)

for(int p=2;p<=k;p++)

//初始分x=b[1]-a[j]

//among分数b[1]是在听了第j个评委报过分之后听到的分数

//cout<<1<<endl;

for(j=1;j<=n;j++)

ans=n;

n=unique(a+1,a+n+1)-a-1;uniquethat函数作用是去重(是将重复的移到后面,分别为b1,b2,...,bn。对于x rated。(每个都不同)当然,每次加完分选手当前得分都会公布一下。然而他只记得其中n个分数(n?≤?k),http://www.xanaxpurchasenow.com/sex_x_ratedhanguonvsheng/20180707/1843.html。然后评委们按次序一个个加上自己给出的分数。已知第i个评委加上了ai分。

sort(a+1,a+n+1);//a[i]sex x rated韩国妈妈是前i个评委打过分之后的分数增量

for(i=1;i<=k;i++) scanf("%d",&b[i]);

for(i=1;i<=n;i++){scanf("%d",&a[i]);a[i]+=a[i-1];}

scanf("%d %d",&n,&k);

//freopen("2.txt","r",stdin);

int main()

int a[2010],b[2010],c[2010];

int n,i,j,k,l,ans;

using namespace std;

#includebits/stdc++.h对比一下he注意这个头文件包括了#includeiostream>#includecstdio>#includefstream>#includealgorithm>#includecmath>#includedeque>#includevector>#includequeue>#includestring>#includecstring>#includemap>#includestack>#includeset>

O(n*k*logn)//--------------------------------------------------------------------------------------

x rated movie下面 若每个b[j]都能找到对应的m 这就是一个可能的答案

所以 不妨设T=b[1] 枚举m 这样得到n个可能的x

不同的T对应不同的m 但每个x保证一个T对应一个m

况且 T的取值是不影响的。因为对一个确定的x

vlaws那么B集合应属于p集合。

p[i]=T-sum[m]+sum[i]

那么Polycarp能听到的分数无外乎

则x=T-sum[m].

如果Polycarpx rated film在第 m 个评委给分时听到了分数 T

记sum[i]= sum(a[1->i]) 则

Polycarp看着wwwJOJO日本能听到的分数就确定下来。 (这里好好想一下)

你看but评委给分是按次序给的。如果初始分数x定下来

b1,b2,...,bn (-≤b≤)//----------------------------------------------------------------------------------

a1,a2,...,ak (-2000我不知道x rated film≤ai≤2000)

k,n(1≤n≤k≤2000)

你的任务就是找出选手的初始分有多少种可能值among。

Your task is to determine the number of options for the scorethe participant could have before the judges rated theparticipant.

Polycarp看着announced不记得选手初始分是多少了。女人有话说。但他记得在评委给分时,即减去)。最初选手有一个初始分,节目中有k个评委给一个选手打分。规则是每个评委给这个选手加上一个自相比看remembers己给出的分值(分值可能为负, Polycarp does not remember how many points the participant hadbefore this k marks were given, but he remembers that among thescores announced after each of the k judges rated the participantthere were n (n?你看remembers≤?k) values b1,b2,..,bn (it is guaranteed that allvalues bj are distinct). It is possible that Polycarp remembers notall of the scores announced, Note that the initial score wasn'tannounced.

Polycarpannounced看了一场节目,日本japds。 Polycarp watched TV-show where k jury members one by one rateda participant by adding him a certain number of points (may benegative, i. e. points were subtracted). Initially the participanthad some score, and each the marks were one by one added to hisscore. It is known that the i-th jury member gave ai points.

announced裁定分数

C. Jury Marks


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