样例输入复制

2
2
3

样例输出复制

1
2

Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only 3030 Yuan).

However, his owner CXY thinks that take-away food is unhealthy and expensive. So she demands her hamster to fulfill a mission before ordering the take-away food. Then she brings the hamster to a wall.

The wall is covered by square ceramic tiles, which can be regarded as a n * mn∗mgrid. CXY wants her hamster to calculate the number of rectangles composed of these tiles.

For example, the following 3 * 33∗3 wall contains 3636 rectangles:

Such problem is quite easy for little hamster to solve, and he quickly manages to get the answer.

Seeing this, the evil girl CXY picks up a brush and paint some tiles into black, claiming that only those rectangles which don't contain any black tiles are valid and the poor hamster should only calculate the number of the valid rectangles. Now the hamster feels the problem is too difficult for him to solve, so he decides to turn to your help. Please help this little hamster solve the problem so that he can enjoy his favorite fried chicken.

Input

There are multiple test cases in the input data.

The first line contains a integer TT : number of test cases. T \le 5T≤5.

For each test case, the first line contains 33integers n , m , kn,m,k , denoting that the wall is a n \times mn×m grid, and the number of the black tiles is kk.

For the next kk lines, each line contains 22integers: x\ yx y ,denoting a black tile is on the xx-th row and yy-th column. It's guaranteed that all the positions of the black tiles are distinct.

For all the test cases,

1 \le n \le 10^5,1\le m \le 1001≤n≤105,1≤m≤100,

0 \le k \le 10^5 , 1 \le x \le n, 1 \le y \le m0≤k≤105,1≤x≤n,1≤y≤m.

It's guaranteed that at most 22 test cases satisfy that n \ge 20000n≥20000.

Output

For each test case, print "Case #xx: ansans" (without quotes) in a single line, where xxis the test case number and ansans is the answer for this test case.

Hint

The second test case looks as follows:

样例输入1复制

5
5 6 0
4 5 1 1
3 4 1 2
2 3 1 3
1 2 1 4

样例输出1复制

55

样例输入2复制

1
-100 0 0

样例输出2复制

0

A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=aband n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1n​f(i).

Input

The first line contains an integer T(T\le 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n(n \le 2\cdot 10^7)n(n≤2⋅107).

Output

For each test case, print the answer \sum_{i = 1}^n f(i)∑i=1n​f(i).

Hint

\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18​f(i)=f(1)+⋯+f(8)
=1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e7+5;
bool pri[N];
int prime[N];
ll f[N];
int tot=0;
int main()
{pri[1]=1,f[1]=1;for(int i=2; i<N; i++){if(pri[i]==0)prime[tot++]=i,f[i]=2;for(int j=0,num; j<tot&&i*1LL*prime[j]<N; ++j){num=i*prime[j],pri[num]=1;if(i%prime[j])f[num]=f[i]*2;else if(i%(1LL*prime[j]*prime[j])==0)f[num]=0;else{f[num]=f[num/prime[j]/prime[j]];break;}}}for(int i=2; i<N; i++)f[i]+=f[i-1];int t,n;scanf("%d",&t);while(t--)scanf("%d",&n),printf("%lld\n",f[n]);
}

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define ll long long
#define sz(x) (int)(x).size()
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pq priority_queue
const int N=2e7+5,MD=1e9+7,INF=0x3f3f3f3f;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
int a[N];
int main()
{ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);for(int i=2; i*i<N; i++)for(int j=i*i; j<N; j+=i*i)a[j]=1;for(int i=1; i<N; i++)a[i]+=a[i-1];int T;cin>>T;while(T--){int n;cin>>n;ll ans=0;int l=1,r;while(l<=n){r=n/(n/l)+1;ans+=1LL*(r-l)*(n/l)-2LL*(a[r-1]-a[l-1])*(n/l)+1LL*(a[r-1]-a[l-1])*a[n/l];l=r;}cout<<ans<<"\n";}return 0;
}

There are NN cities in the country, and MMdirectional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become 00. Now she wants to go to City NN, please help her calculate the minimum distance.

Input

The first line has one integer T(1 \le T\le 5)T(1≤T≤5), then following TT cases.

For each test case, the first line has three integers N, MN,M and KK.

Then the following MM lines each line has three integers, describe a road, U_i, V_i, C_iUi​,Vi​,Ci​. There might be multiple edges between uu and vv.

It is guaranteed that N \le 100000, M \le 200000, K \le 10N≤100000,M≤200000,K≤10,
0 \le C_i \le 1e90≤Ci​≤1e9. There is at least one path between City 11 and City NN.

Output

For each test case, print the minimum distance.

#include<bits/stdc++.h>
using namespace std;#define ll long long
const int maxn=1e5+5,maxm=2e5+5;
ll d[11][maxn];
int n,m,k;
int head[maxn],cnt;
struct edge
{int v,next;ll w;}edges[maxm];
struct node
{ll w;int v;bool operator<(const node &D)const{return w>D.w;}
};
inline bool read(int &num)
{char in;bool IsN=false;in=getchar();if(in==EOF) return false;while(in!='-'&&(in<'0'||in>'9')) in=getchar();if(in=='-'){IsN=true;num=0;}else num=in-'0';while(in=getchar(),in>='0'&&in<='9'){num*=10,num+=in-'0';}if(IsN) num=-num;return true;
}
void dij()
{memset(d,0x3f,sizeof d);priority_queue<node>q;q.push({0,1});d[0][1]=0;while(!q.empty()){auto u=q.top();q.pop();if(u.v==n)return;for(int i=head[u.v];i!=-1;i=edges[i].next){edge &v=edges[i];for(int j=0;j<=k;j++){if(j>=1&&d[j][v.v]>d[j-1][u.v])q.push({d[j][v.v]=d[j-1][u.v],v.v});if(d[j][v.v]>d[j][u.v]+v.w)q.push({d[j][v.v]=d[j][u.v]+v.w,v.v});}}}
}
int main()
{int t;read(t);while(t--){memset(head,-1,sizeof head);cnt=0;read(n),read(m),read(k);for(int i=0,u,v,w;i<m;i++){read(u);read(v);read(w);edges[cnt].v=v;edges[cnt].w=1LL*w;edges[cnt].next=head[u];head[u]=cnt++;}dij();ll minn=0x3f3f3f3f3f3f3f3f;for(int i=0;i<=k;i++)minn=min(minn,d[i][n]);printf("%lld\n",minn);}return 0;
}