结果：2350/9339
C挂了，恶心人啊，cf电脑一直登不上，还是手机查的，这场比赛好气啊；
A. Display The Number
解法：若n为偶数，输出n/2个1；若n为奇数，输出7和（n-3)/2个1.
WA点：7放后面WA了一发。
ACODE:
/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define fill(x,c) memset(x,c,sizeof(x))
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
const int SIZE = 1e6+10;
ll a[SIZE];
int main() {
//ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t,n;
cin>>t;
while(t--)
{
cin>>n;
if(n&1){
int tmp = n-3;
cout<<7<<string(tmp/2,'1')<<endl;
}else{
cout<<string(n/2,'1')<<endl;
}
}
return 0;
}
Infinite Prefixes
解法：定义pre[i]表示长度为i的前缀平衡值，最后得到的前缀为k（k>=0) 个完整串和长度为i(0~n-1)的字串组成，遍历i从0至n-1，分情况讨论：
WA点：pre[n] = abs(pre[n]) 产生副作用WA了2发。
ACODE:
/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define fill(x,c) memset(x,c,sizeof(x))
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
const int SIZE = 1e6+10;
ll pre[SIZE];
int main() {
//ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
ll t,n,x,cnt0,cnt1;
string s;
cin>>t;
while(t--){
cin>>n>>x>>s;
cnt0=0,cnt1=0;
fill(pre,0);
rep(i,0,n){
pre[i] = cnt0-cnt1;
(s[i]=='0'?cnt0++:cnt1++);
}
pre[n] = cnt0-cnt1;
ll f = 0,ans = 0;
ll N = pre[n] ;
rep(i,0,n){
ll left = x-pre[i];
pre[n] = N;
if(left*pre[n]<0){
;
}else{
left = abs(left);
pre[n] = abs(pre[n]);
if(pre[n]==0){
if(left==0){
cout<<-1<<endl;
f = 1;
break;
}else{
;
}
}else{
if(left%pre[n]==0){
ans++;
}
}
}
}
if(f)continue;
cout<<ans<<endl;
}
return 0;
}
Obtain The String
解法：保存每个字符在s中的位置数组，遍历字符串t，看当前字符在前一字符的位置后面是否存在。
WA点：没有用while找位置，而是单纯从0开始判断。
ACODE：
/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define fill(x,c) memset(x,c,sizeof(x))
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
const int SIZE = 1e6+10;
int a[26];
int b[26];
int c[26];
int main() {
//ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T,n,m,f;
string s,t;
cin>>T;
while(T--){
cin>>s>>t;
fill(a,0);
fill(b,0);
fill(c,0);
unordered_map<char,vector<int> > P;
n = s.size(),m = t.size();
f = 0;
rep(i,0,n){
a[s[i]-'a']++;
P[s[i]].pb(i);
}
rep(i,0,m){
if(a[t[i]-'a']==0){
f=1;
cout<<-1<<endl;
break;
}
}
if(f)continue;
int i=0,ans=1;
int pre = -1;
while(i<m){
while(c[t[i]-'a']<(int)P[t[i]].size()&&P[t[i]][c[t[i]-'a']]<=pre)c[t[i]-'a']++;
if(c[t[i]-'a']<(int)P[t[i]].size()){
pre = P[t[i]][c[t[i]-'a']];
c[t[i]-'a']++;
i++;
}else{
ans++;
fill(c,0);
pre = -1;
}
}
cout<<ans<<endl;
}
return 0;
}
来源：CSDN
作者：free_steed
链接：https://blog.csdn.net/Csdn_jey/article/details/104111722