Little Boxes
Problem Description
Little boxes on the hillside.
Little boxes made of ticky-tacky.
Little boxes.
Little boxes.
Little boxes all the same.
There are a green boxes, and b pink boxes.
And c blue boxes and d yellow boxes.
And they are all made out of ticky-tacky.
And they all look just the same.
Little boxes made of ticky-tacky.
Little boxes.
Little boxes.
Little boxes all the same.
There are a green boxes, and b pink boxes.
And c blue boxes and d yellow boxes.
And they are all made out of ticky-tacky.
And they all look just the same.
Input
The input has several test cases. The first line contains the integer t (1 ≤ t ≤ 10) which is the total number of test cases.
For each test case, a line contains four non-negative integers a, b, c and d where a, b, c, d ≤ 2^62, indicating the numbers of green boxes, pink boxes, blue boxes and yellow boxes.
For each test case, a line contains four non-negative integers a, b, c and d where a, b, c, d ≤ 2^62, indicating the numbers of green boxes, pink boxes, blue boxes and yellow boxes.
Output
For each test case, output a line with the total number of boxes.
Sample Input
4 1 2 3 4 0 0 0 0 1 0 0 0 111 222 333 404
Sample Output
10 0 1 1070
Java 即可；
import java.math.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
BigInteger a,b,c,d;
int T;T=cin.nextInt();
for(int i=0;i<T;i++) {
a=cin.nextBigInteger();b=cin.nextBigInteger();c=cin.nextBigInteger();d=cin.nextBigInteger();
BigInteger res=a.add(b).add(c).add(d);
System.out.println(res);
}
}
}
Rabbits
Problem Description
Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number line, each occupying a different integer. In a single move, one of the outer rabbits jumps into a space between any other two. At no point may two rabbits occupy the same position.
Help them play as long as possible
Help them play as long as possible
Input
The input has several test cases. The first line of input contains an integer t (1 ≤ t ≤ 500) indicating the number of test cases.
For each case the first line contains the integer N (3 ≤ N ≤ 500) described as above. The second line contains n integers a1 < a2 < a3 < ... < aN which are the initial positions of the rabbits. For each rabbit, its initial position
ai satisfies 1 ≤ ai ≤ 10000.
For each case the first line contains the integer N (3 ≤ N ≤ 500) described as above. The second line contains n integers a1 < a2 < a3 < ... < aN which are the initial positions of the rabbits. For each rabbit, its initial position
ai satisfies 1 ≤ ai ≤ 10000.
Output
For each case, output the largest number of moves the rabbits can make.
Sample Input
5 3 3 4 6 3 2 3 5 3 3 5 9 4 1 2 3 4 4 1 2 4 5
Sample Output
1 1 3 0 1
前后分别扫一遍即可，取max；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
/*
int n, m;
int st, ed;
struct node {
int u, v, nxt, w;
}edge[maxn<<1];
int head[maxn], cnt;
void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w;
edge[cnt].nxt = head[u]; head[u] = cnt++;
}
int rk[maxn];
int bfs() {
queue<int>q;
ms(rk);
rk[st] = 1; q.push(st);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].v;
if (rk[to] || edge[i].w <= 0)continue;
rk[to] = rk[tmp] + 1; q.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(edge[i].w, flow - add));
if (!tmpadd) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
}
return add;
}
ll ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
*/
int n;
int a[1000];
int main()
{
//ios::sync_with_stdio(0);
//memset(head, -1, sizeof(head));
int T; cin >> T;
while (T--) {
rdint(n);
for (int i = 1; i <= n; i++)rdint(a[i]);
int ans = 0;
for (int i = 2; i < n; i++) {
ans += (a[i + 1] - a[i] - 1);
}
int maxx = ans;
ans = 0;
for (int i = n - 1; i > 1; i--) {
ans += (a[i] - a[i - 1] - 1);
}
maxx = max(maxx, ans);
cout << maxx << endl;
}
return 0;
}
Heron and His Triangle
Problem Description
A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t−1, t, t+ 1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger
than or equal to n.
than or equal to n.
Input
The input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30).
Output
For each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.
Sample Input
4 1 2 3 4
Sample Output
4 4 4 4
找规律：先打表出前几项；
可以发现 num[ i ]=num[ i-1 ]*4-num[ i-2 ];
然后大数即可；
import java.lang.reflect.Array;
import java.math.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
BigInteger N;
int T;T=cin.nextInt();
BigInteger num[]=new BigInteger[1000];
num[1]=BigInteger.valueOf(4);num[2]=BigInteger.valueOf(14);
for(int i=3;i<=500;i++)num[i]=num[i-1].multiply(BigInteger.valueOf(4)).subtract(num[i-2]);
for(int i=0;i<T;i++) {
N=cin.nextBigInteger();
for(int j=1;j<=400;j++) {
if(num[j].compareTo(N)>=0) {
System.out.println(num[j]);break;
}
}
}
}
}
Tree
Problem Description
Consider a un-rooted tree T which is not the biological significance of tree or plant, but a tree as an undirected graph in graph theory with n nodes, labelled from 1 to n. If you cannot understand the concept of a tree here, please omit this problem.
Now we decide to colour its nodes with k distinct colours, labelled from 1 to k. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset of edges connecting all nodes coloured by i. If there is no node of the tree coloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, and output its size.
Now we decide to colour its nodes with k distinct colours, labelled from 1 to k. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset of edges connecting all nodes coloured by i. If there is no node of the tree coloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, and output its size.
Input
The first line of input contains an integer T (1 ≤ T ≤ 1000), indicating the total number of test cases.
For each case, the first line contains two positive integers n which is the size of the tree and k (k ≤ 500) which is the number of colours. Each of the following n - 1 lines contains two integers x and y describing an edge between them. We are sure that the given graph is a tree.
The summation of n in input is smaller than or equal to 200000.
For each case, the first line contains two positive integers n which is the size of the tree and k (k ≤ 500) which is the number of colours. Each of the following n - 1 lines contains two integers x and y describing an edge between them. We are sure that the given graph is a tree.
The summation of n in input is smaller than or equal to 200000.
Output
For each test case, output the maximum size of E1 ∩ E1 ... ∩ Ek.
Sample Input
3 4 2 1 2 2 3 3 4 4 2 1 2 1 3 1 4 6 3 1 2 2 3 3 4 3 5 6 2
Sample Output
1 0 1
题意：
考虑一个非根树，它不是树或植物的生物学意义，而是一个树作为图论中的无向图，有n个节点，从1到n标记。 如果你在这里无法理解树的概念，请省略这个问题。
现在我们决定用k个不同的颜色为节点着色，标记为1到k。 然后，对于每种颜色i = 1,2，...，k，将Ei定义为连接由i着色的所有节点的边的最小子集。 如果树的节点没有由指定颜色i着色，则Ei将为空。
尝试确定一个配色方案，以最大化E1∩E2···∩Ek的大小，并输出其大小。
现在我们决定用k个不同的颜色为节点着色，标记为1到k。 然后，对于每种颜色i = 1,2，...，k，将Ei定义为连接由i着色的所有节点的边的最小子集。 如果树的节点没有由指定颜色i着色，则Ei将为空。
尝试确定一个配色方案，以最大化E1∩E2···∩Ek的大小，并输出其大小。
很明显的一点是：我们K种颜色都要使用，否则一个集合为0，则其交集必然为0；
不仅如此，当我们割去一条边时，树分为两个部分，若两部分大小均>=K，那么ans++;
因为我们枚举的是边，那么如果是交集里的边，则两边的numV>=k,那么我们dfs一次求其子树大小即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
/*
int n, m;
int st, ed;
struct node {
int u, v, nxt, w;
}edge[maxn<<1];
int head[maxn], cnt;
void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w;
edge[cnt].nxt = head[u]; head[u] = cnt++;
}
int rk[maxn];
int bfs() {
queue<int>q;
ms(rk);
rk[st] = 1; q.push(st);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].v;
if (rk[to] || edge[i].w <= 0)continue;
rk[to] = rk[tmp] + 1; q.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(edge[i].w, flow - add));
if (!tmpadd) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
}
return add;
}
ll ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
*/
int n;
vector<int>vc[maxn];
int Siz[maxn];
void dfs(int u, int rt) {
Siz[u] = 1;
for (int i = 0; i < vc[u].size(); i++) {
int v = vc[u][i];
if (v == rt)continue;
dfs(v, u); Siz[u] += Siz[v];
}
}
int main()
{
//ios::sync_with_stdio(0);
//memset(head, -1, sizeof(head));
int T; rdint(T);
while (T--) {
int k; rdint(n); rdint(k);
for (int i = 0; i < maxn; i++)vc[i].clear();
for (int i = 1; i < n; i++) {
int u, v; rdint(u); rdint(v);
vc[u].push_back(v); vc[v].push_back(u);
}
dfs(1, 0);
int ans = 0;
for (int i = 1; i <= n; i++) {
if (Siz[i] >= k && (n - Siz[i]) >= k)ans++;
}
cout << ans << endl;
}
return 0;
}