A
/*
* Author: Moon-light
* 按照題意輸出
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 1;
const int mod = 1e9+7;
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
//
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<min(a,min(b,min(c,d)))<<endl;
return 0;
}
B
/*
* Author: Moon-light
* 按照題意邊模擬邊判斷即可
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 1;
const int mod = 1e9+7;
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
//
LL m,n,t;
LL a,b;
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
cin>>n>>m>>t;
LL left = n;
LL last = 0;
m -= 1;
cin>>a>>b;
left -= a;
if(left <= 0) {
cout<<"No"<<endl;
return 0;
}
left += (b-a);
last = b;
if(left > n) left = n;
for(int i = 1; i<=m; i++){
cin>>a>>b;
left = left - (a-last);
if(left <= 0) {
cout<<"No"<<endl;
return 0;
}
left = left + (b-a);
if(left > n) left = n;
last = b;
}
left = left - (t-last);
if(left <= 0){
cout<<"No"<<endl;
}else{
cout<<"Yes"<<endl;
}
return 0;
}
C
/*
* Author: Moon-light
* 題意：一根木棍分成12段的切法數(shù)量，兩種切法不同當(dāng)且僅當(dāng)切割的位置不同
* 求組合數(shù)即可士修，注意一個(gè)木棍可切割的位置是n-1
* 那么answer = C(11,n-1);
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 300;
const int mod = 1e9+7;
LL c[N][N];
int n;
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
void solve(){
for(int i = 0; i<N; i++){
for(int j = 0; j<=i; j++){
if(!j) c[i][j] = 1;
else c[i][j] = c[i-1][j] + c[i-1][j-1];
}
}
}
//
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
cin>>n;
solve();
cout<<c[n-1][11]<<endl;
return 0;
}
D
/*
* Author: Moon-light
*
* 思路：因?yàn)椴荒芘龅剿{(lán)色的方塊，所以我們只能找藍(lán)色方塊分割開的白色序列中找最小的連續(xù)的子序列
* 然后求次數(shù)即可，注意首尾的邊界情況
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 2e5+100;
const int mod = 1e9+7;
LL n,m;
LL a[N];
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
//
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
cin>>n>>m;
for(int i = 1; i<=m; i++){
cin>>a[i];
}
sort(a+1,a+1+m);
if(m == 0){
cout<<1<<endl;
return 0;
}
if(n == m) {
cout<<0<<endl;
return 0;
}
LL range = a[1] - 1;
LL last = a[1];
for(int i = 2; i<=m; i++){
LL t = a[i] - last - 1;
if(range == 0) range = t;
else if(t > 0) range = min(range,t);
last = a[i];
}
if(n > last){
if(range == 0) range = (n-last);
else range = min(range,n-last);
}
//cout<<range<<endl;
LL ans = 0;
last = a[1];
if(a[1] > 1){
ans += (a[1]-2)/range+1;
}
for(int i = 2; i<=m; i++){
int cnt = a[i]-last-1;
ans += (cnt-1)/range+1;
last = a[i];
}
if(n>last)
ans += (n-last-1)/range + 1;
cout<<ans<<endl;
return 0;
}
E
/*
* Author: Moon-light
*
* 題意: 有兩個(gè)長度分別為N和M的序列A,B
* 刪除序列A中的a個(gè)元素得到A' , 0<=a<=N
* 刪除序列B中的b個(gè)元素得到B' , 0<=b<=M
* 使得A'的長度等于B'的長度
* 令x = a + b
* y = Sum of i A'i != B'i
*
* 思路: 編輯距離模型
* 設(shè)dp[i][j] 表示對(duì)A前i個(gè)及對(duì)B前j個(gè)元素的最小答案
* 那么對(duì)于當(dāng)前的A[i]和 B[j]
* 有三種情況： 1.刪除A[i]，保留B[j] dp[i-1][j] + 1
* 2.保留A[i]，刪除B[j] dp[i][j-1] + 1
* 3.保留A[i], 保留A[j] dp[i-1][j-1] + (a[i]==b[j]?0:1)
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 1010;
const int mod = 1e9+7;
int n,m;
int a[N],b[N];
int dp[N][N];
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
//
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
cin>>n>>m;
rep(i,1,n) cin>>a[i];
rep(i,1,m) cin>>b[i];
//initialization
rep(i,1,n) dp[i][0] = i;
rep(i,1,m) dp[0][i] = i;
rep(i,1,n){
rep(j,1,m){
//刪一個(gè)
dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1);
//不刪
dp[i][j] = min(dp[i][j], dp[i-1][j-1] + (a[i]==b[j]?0:1));
}
}
cout<<dp[n][m];
return 0;
}
F
/*
* Author: Moon-light
* 題意: 1.單點(diǎn)修改 Ax = Ax ^ Y
* 2.區(qū)間查詢 Ax1 ^ Ax2 ^ Ax3 ^... ^ Ay
*
* 思路: 樹狀數(shù)組模板
*/
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define lowbit(x) ((x)&(-x))
#define sz(x) ((int)x.size())
#define fr(x) freopen(x,'r',stdin)
#define fw(x) freopen(x,'w',stdout)
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define rep(i,a,b) for(int i = (a);i<=(b); ++i)
#define per(i,a,b) for(int i = (a);i>=(b); --i)
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<double,double> PDD;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<double> VB;
typedef double db;
template <class T> void clear(T& a) { T().swap(a);}
template <class T,class S>
inline bool upmin(T& a,const S&b){ return a>b ? a=b,1:0;}
template <class T,class S>
inline bool upmax(T&a, const S&b){ return a<b? a=b,1:0;}
//fast read(improved by fread)
char buf[1<<21],*p1=buf,*p2=buf;
inline int getc(){
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
}
inline int redi() {
int ret = 0,f = 0;char ch = getc();
while (!isdigit (ch)) {
if (ch == '-') f = 1;
ch = getc();
}
while (isdigit (ch)) {
ret = ret * 10 + ch - 48;
ch = getc();
}
return f?-ret:ret;
}
//global variable
const int N = 3e5+100;
const int mod = 1e9+7;
int n,q,t,x,y;
int tr[N];
//functions
LL qmi(LL a,LL k,LL mod){
LL res = 1;
while(k){
if(k&1) res = res*a%mod;
k >>= 1;
a = a*a%mod;
}
return mod;
}
int gcd(int a,int b) {
return b==0? a:gcd(b,a%b);
}
void update(int u,int x){
for(int i = u; i<=n; i+=lowbit(i)){
tr[i] ^= x;
}
}
int query(int x){
int res = 0;
for(int i = x; i; i-=lowbit(i)){
res ^= tr[i];
}
return res;
}
//
//code from here! Come on! Have a pleasant experience~
int main()
{
IO;
cin>>n>>q;
rep(i,1,n){
cin>>x;
update(i,x);
}
rep(i,1,q){
cin>>t>>x>>y;
if(t == 1){
update(x,y);
}else{
cout<<(query(y)^query(x-1))<<endl;
}
}
return 0;
}
