A Erasing Zeroes
直接模拟
B National Project
算出最后一轮前需要几轮，然后加上剩余需要的天数。最后判断一下是不是比n小，如果是就再多做几天。
这题wa 3发心态小崩
#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
键盘前两个字母根据密码确定之后后面的就都确定了。
#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
判断一下Sum是不是小于n，小于n肯定无解，否则一定有解（全部换成1）
然后从n的低位开始凑，凑完低位之后这一位剩下的都可以往上合成。如果这一位没有，就往上找，从上面翻下来。因为输入的 a i a_i ai小于1e9所以只要做到30位就足够了。
#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
枚举分裂点，分成两个串。然后 d p ( i , j ) dp(i,j) dp(i,j)表示第一串匹配到 i i i，第二串匹配到 j j j至少需要匹配到 s s s的第几个位置去判断。
#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;
}
}
/*
1
defi
fed
*/