AtCoder Beginner Contest 167
abc过的挺顺利的到D就卡了，又wa又T
A.Registration
题目大意
题目意思大概就是给你字符串S和T，确保T的长度比S多1，求S是否是T的前缀
思路
暴力模拟
代码
/*********************
*@Author: CKang *
*@Language: C++11 *
*********************/
#include<bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
//const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-9;
const int INF = 0x3fffffff;
const ll LINF = 0x3fffffffffffffff;
inline ll read(){
long long x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);
ch=getchar();
}
return x*f;
}
void solve(){
string s,t;
cin>>s>>t;
for(int i=0;i<s.length();i++){
if(s[i]!=t[i]){
puts("No");
return ;
}
}
puts("Yes");
return ;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout)
#endif
//cout.tie(0);
//for(ll t=read();t;t--)
solve();
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Fuck You !" << endl;
return 0;
}
B.Easy Linear Programming
题目大意
给你a个1，b个0，c个-1，共a+b+c个数字以及一个整数k，求从这些数字中选k个求和，要求和最大
思路
先拿1再是0再是-1
代码
/*********************
*@Author: CKang *
*@Language: C++11 *
*********************/
#include<bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
//const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-9;
const int INF = 0x3fffffff;
const ll LINF = 0x3fffffffffffffff;
inline ll read(){
long long x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);
ch=getchar();
}
return x*f;
}
void solve(){
ll a=read(),b=read(),c=read(),k=read(),sum=0;
if(k>a){
sum+=a;
k-=a;
if(k>b){
k-=b;
if(k>c){
return ;
}
else {
printf("%d\n",sum-k);
}
}
else {
printf("%d\n",sum);
}
}
else {
printf("%d\n",k);
}
return ;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout)
#endif
//cout.tie(0);
//for(ll t=read();t;t--)
solve();
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Fuck You !" << endl;
return 0;
}
C.Skill Up
题目大意
对于n本书，每一本书都有价值c，以及对m个不同能力的不同的增益，求各个增益均大于x时所挑选书的c之和最小值
思路
每本书都是选和不选，数据小直接dfs了，数据大一点就要dp了（但是不会）
代码
//********************
//*@Author: ChenShou*
//*@Language:C++11 *
//********************
#include<bits/stdc++.h>
//#pragma comment ( linker ," /STACK:102400000 , 102400000 " )
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std ;
typedef long long ll ;
// typedef __int128 lll ;
const int MOD = 1e9 + 7 ;
const double PI = acos(-1.0) ;
const double EXP = 1E-9 ;
const int INF = 0x3f3f3f3f;
inline ll read(){
long long x = 0 , f = 1 ;
char ch = getchar() ;
while(ch<'0'||ch>'9'){
if(ch == '-') f = -1 ;
ch = getchar();
}
while(ch >='0' && ch <= '9' ){
x = (x<<1) + (x<<3) + (ch^48);
ch = getchar();
}
return x*f ;
}
struct mas{
int c;
int a[15];
};
mas sum={0};
ll N,M,X,min_c=INF;
mas mass[15];
void dfs(int n,mas tmp){
if(n==N)return ;
bool flag=true;
dfs(n+1,tmp);
tmp.c+=mass[n].c;
for(int i=0;i<M;i++){
tmp.a[i]+=mass[n].a[i];
if(tmp.a[i]<X){
flag=false;
}
}
if(flag&&tmp.c<min_c)min_c=tmp.c;
dfs(n+1,tmp);
return ;
}
void solve(){
N=read(),M=read(),X=read();
for(int i=0;i<N;i++){
mass[i].c=read();
for(int j=0;j<M;j++){
mass[i].a[j]=read();
}
}
dfs(0,sum);
if(min_c!=INF)cout<<min_c<<endl;
else cout<<-1<<endl;
return ;
}
int main(){
#ifdef DEBUG
freopen("input.in","r",stdin);
freopen("output.out","w",stdout);
#endif
// for(ll T=read();T;T--)
solve();
#ifdef DEBUG
printf("\nHellow , Mr.ChenShou\n");
printf("\t Time cost : %lf s \n",(double)clock()/CLOCKS_PER_SEC);
#endif
return 0;
}
D.Teleporter
题目大意
题目意思大概就是每一个城市可以传送到另一个城市，从1号城市开始传送k次到那个城市
思路
对于次数多的时候必有环，直接标记走过的点走过几次，只走过一次的点就是进入环之前的铺垫，走过两次的就是环上的点了，记录环的长度后求余即可。
代码
/*********************
*@Author: CKang *
*@Language: C++11 *
*********************/
#include<bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
//const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-9;
const int INF = 0x3fffffff;
const ll LINF = 0x3fffffffffffffff;
inline ll read(){
long long x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);
ch=getchar();
}
return x*f;
}
int nums[200005]={0};int mp[200005]={0};
void solve(){
ll n=read(),k=read(),city=1,cir=0,fir;
for(int i=1;i<=n;i++)nums[i]=read();
for(;;){
mp[city]++;
if(mp[city]==3){
fir=city;
break;
}
if(mp[city]==2){
cir++;
}
if(mp[city]==1){
k--;
if(k==0){ //就是这里忘记判断了，又T又wa
cout<<nums[city]<<endl;
return ;
}
}
city=nums[city];
}
k=k%cir;
while(k){
fir=nums[fir];
k--;
}
cout<<fir<<endl;
return ;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout)
#endif
//cout.tie(0);
//for(ll t=read();t;t--)
solve();
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Fuck You !" << endl;
return 0;
}
