A Erasing Zeroes
Direct simulation
B National Project
Calculate the number of rounds required before the last round, and then add the remaining days. Finally, judge if it is smaller than n, and if so, do it for a few more days.
This question wa 3 times the mentality collapsed
#include<bits/stdc++.h>
using namespace std;
int main()
{
int T;cin>>T;
while(T--){
ll n, g, b;
cin>>n>>g>>b;
ll a = n;
n = (n+1)/2;
if(n == 0) {
cout<<a<<endl; continue;
}
ll ans = (n-1)/g * (g+b);
ans += n-(n-1)/g*g;
if(ans < a) ans = a;
cout<<ans<<endl;
}
}
C Perfect Keyboard
After the first two letters of the keyboard are confirmed according to the password, the latter are confirmed.
#include<bits/stdc++.h>
using namespace std;
int vis[26];
int main()
{
int T;cin>>T;
while(T--){
string s; cin>>s;
string ans; ans.clear();
memset(vis, 0, sizeof vis);
ans += s[0]; int c = 0;
vis[s[0]-'a'] = 1;
int f = 1;
for(int i = 1; i < s.size(); ++i){
int x = s[i]-'a';
//cout<<"x:"<<x<<endl;
if(c+1 == ans.size() && !vis[x]){
ans += s[i]; c++; vis[x] = 1;
}else if(!vis[x]){
vis[x] = 1;
if(c == 0) ans = (char)(s[i])+ans;
else {
f = 0; break;}
}else{
if(c+1 < ans.size() && ans[c+1] == s[i]) c++;
else if(c-1 >= 0 && ans[c-1] == s[i]) c--;
else {
f = 0; break;}
}
}
if(!f) cout<<"NO"<<endl;
else {
cout<<"YES"<<endl;
for(int i = 0; i < 26; ++i) if(!vis[i]) ans += (char)(i+'a');
cout<<ans<<endl;
}
}
}
D Fill The Bag
Judge whether Sum is less than n, and there must be no solution if it is less than n, otherwise there must be a solution (all replaced by 1)
and then start from the low bit of n. After the low bit, the rest of this bit can be combined up. If this one doesn't exist, look up and flip down from above. Because the input ai a_iaiIt is less than 1e9, so only 30 bits are enough.
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
int vis[64];
int main()
{
int T;cin>>T;
while(T--){
memset(vis,0,sizeof vis);
ll n; cin>>n;
int m; cin>>m;
ll sum = 0;
while(m--){
int x; scanf("%d", &x);
for(int i = 0; i < 30; ++i) if(x>>i&1) {
vis[i]++; break;}
sum += x;
}
if(sum < n){
cout<<-1<<endl; continue;
}
int ans = 0;
for(int i = 0; i < 32; ++i){
if(n>>i&1){
if(vis[i]) vis[i]--;
else {
for(int j = i+1; j < 32; ++j){
if(vis[j]){
while(j > i) {
vis[j]--;
j--; vis[j] += 2;
ans++;
}
break;
}
}
vis[i]--;
}
}
vis[i+1]+=(vis[i]/2);
}
cout<<ans<<endl;
}
}
E Erase Subsequences
Enumerate split points and divide them into two strings. Then dp (i, j) dp(i,j)dp(i,j ) means that the first string matchesiii , the second string matchesjjj needs to match at leastsss of several locations to judge.
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
const int maxn = 405;
int nxt[405][26];
char s[maxn];
char t[maxn];
int n;
int dp[maxn][maxn];
bool go(int n1, char *a, int n2, char *b){
dp[0][0] = 0;
for(int i = 0; i <= n1; ++i){
for(int j = 0; j <= n2; ++j){
if(i+j == 0) continue;
dp[i][j] = n+1;
if(i) dp[i][j] = min(dp[i][j], nxt[dp[i-1][j]][a[i]-'a']);
if(j) dp[i][j] = min(dp[i][j], nxt[dp[i][j-1]][b[j]-'a']);
}
}
return dp[n1][n2] <= n;
}
int main()
{
int T; cin>>T;
while(T--){
scanf("%s", s+1);
scanf("%s", t+1);
n = strlen(s+1);
for(int i = 0; i < 26; ++i) nxt[n][i] = nxt[n+1][i] = n+1;
for(int i = n-1;i >= 0;--i){
for(int j = 0;j < 26;++j) nxt[i][j] = nxt[i+1][j];
nxt[i][s[i+1]-'a'] = i + 1;
}
int f = 1;
int p = 0;
int len = strlen(t+1);
for(int i = 1; i <= len; ++i){
int x = t[i]-'a';
p = nxt[p][x];
if(p == n+1) f = 0;
}
if(f){
cout<<"YES"<<endl; continue;
}
for(int i = 1; i < len; ++i){
if(f) break;
if(go(i, t, len-i, t+i)){
f = 1;
}
}
if(f) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
/*
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defi
fed
*/