A.String
Transmission: https://ac.nowcoder.com/acm/contest/887/A
The meaning of problems: $ 0, $ a given string, the minimum requirements for dividing portion, such that each portion is lexicographic lexicographically smallest cycle.
Data range: $ 1 <= T <= 300,1 <= len <= 200 $.
Analysis: Violence to split into a zero string ends pairwise merge, not until combined.
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=300; 4 string aa[maxn],bb[maxn]; 5 char s[maxn]; 6 int tot,tot2; 7 void init(){ 8 int len=strlen(s+1); 9 string ss=""; 10 int flag=1; tot=0; 11 for (int i=1;i<=len;i++){ 12 if(s[i]=='1') ss+=s[i],flag=0; 13 else if(s[i]=='0'){ 14 if(flag) ss+=s[i]; 15 else{ 16 aa[++tot]=ss; ss=s[i]; flag=1; 17 } 18 } 19 if(i==len) aa[++tot]=ss; 20 } 21 } 22 bool check(){ 23 for (int i=1;i<tot;i++){ 24 string x=aa[i],y=aa[i+1]; 25 if(x+y<=y+x) return true; 26 } 27 return false; 28 } 29 string st[maxn]; 30 void solve(){ 31 int top=1; st[1]=""; 32 for (int i=1;i<=tot;i++){ 33 string a=st[top]; 34 string b=aa[i]; 35 string tmp1=a+b,tmp2=b+a; 36 if(tmp1<=tmp2) st[top]=tmp1; 37 else st[++top]=b; 38 } 39 tot=0; 40 for (int i=1;i<=top;i++) aa[++tot]=st[i]; 41 } 42 int main(){ 43 int t;scanf("%d",&t); 44 while (t--){ 45 scanf("%s",s+1); 46 init(); 47 while (1){ 48 if (check()) solve(); 49 else break; 50 } 51 for (int i=1;i<=tot;i++) 52 if (i==1) cout << aa[i]; 53 else cout << " " << aa[i]; 54 cout << endl; 55 } 56 return 0; 57 }
B.Irreducible Polynomial
Transmission: https://ac.nowcoder.com/acm/contest/887/B
Meaning of the questions: Given a polynomial, can be asked about whether.
Data range: $ 1 <= T <= 100,1 <= n <= 20, -10 ^ 9 <= a_i <= 10 ^ 9 $.
Analysis: real domain can not be split only two polynomials: a quadratic polynomial and ($ b ^ 2 <4ac $).
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 ll a[100]; 5 int main() 6 { 7 int t;scanf("%d",&t); 8 while (t--) 9 { 10 int n;scanf("%d",&n); 11 for (int i=n;i>=0;i--) scanf("%lld",&a[i]); 12 if(n==0 || n==1) {printf("Yes\n");continue;} 13 if(n>=3) {printf("No\n");continue;} 14 ll dd=a[1]*a[1]-4*a[0]*a[2]; 15 if(dd<0) printf("Yes\n"); 16 else printf("No\n"); 17 } 18 return 0; 19 }
C.Governing sand
Transmission: https://ac.nowcoder.com/acm/contest/887/C
Meaning of the questions: There are $ n $ trees, trees have a high degree per $ h_i $, cut spending $ c_i $, the number of $ p_i $. Highest number of trees must be> = half, and asked other trees smallest cut costs as much.
Data range: $ 1 <= T <= 30,1 <= n <= 10 ^ 5,1 <= h_i, p_i <= 10 ^ 9,1 <= c_i <= 200 $.
Analysis: From low to the highest high enumeration tree, the rest of the tree which cut the cost minimal. With weights segment tree maintenance.
Segment tree according to the cost of building from small to large, and then query cost as low as possible.
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #define m(a,b) memset(a,b,sizeof a) 6 #define en '\n' 7 using namespace std; 8 typedef long long ll; 9 const int N=2e5+5; 10 const ll INF=1ll<<61ll; 11 struct node{int id;ll h;}a[N]; 12 int cmp(node x,node y){return x.h<y.h;} 13 ll c[N],num[N],sum[N],sb[N];int r[N]; 14 ll tag[N<<2],tsum[N<<2]; 15 void build(ll l,ll r,ll pos){ 16 tag[pos]=tsum[pos]=0; 17 if(l==r) return; 18 ll mid=(l+r)>>1; 19 build(l,mid,pos<<1); 20 build(mid+1,r,pos<<1|1); 21 } 22 void update(ll dex,ll num,ll l,ll r,ll pos) 23 { 24 if(l==r) { 25 tag[pos]+=num,tsum[pos]+=num*l; 26 return; 27 } 28 ll mid=(l+r)>>1; 29 if(dex<=mid) update(dex,num,l,mid,pos<<1); 30 else update(dex,num,mid+1,r,pos<<1|1); 31 tag[pos]=tag[pos<<1]+tag[pos<<1|1];//左边的数量 32 Tsum [POS] = Tsum [<< POS . 1 ] Tsum + [POS << . 1 | . 1 ]; // left money 33 is } 34 is LL Query (cnt LL, LL L, R & lt LL, LL POS) // find cnt a minimum cost of 35 { 36 IF (R & lt == L) return L * CNT; 37 [ LL MID = (L + R & lt) >> . 1 ; 38 is IF (Tag [<< POS . 1 ]> = CNT) return Query (CNT , L, MID, << POS . 1 ); // number be left 39 the else return Tsum [<< POS . 1 ] + Query (CNT-Tag [<< POS . 1],mid+1,r,pos<<1|1);//左边的金钱+右边询问 40 } 41 int main(){ 42 int n; 43 while(~scanf("%d",&n)) 44 { 45 ll _A=0,Max=0; 46 for(int i=1;i<=n;++i) 47 scanf("%lld%lld%lld",&a[i].h,&c[i],&num[i]),a[i].id=i,_A+=num[i],Max=max(Max,c[i]); 48 build(1,Max,1); 49 sort(a+1,a+n+1,cmp); a[n+1].h=0; 50 for (int i=n;i>=1;i--) 51 { 52 if (a[i].h==a[i+1].h) r[i]=r[i+1]; 53 else r[i]=i; 54 } 55 ll res=0,res2=0; 56 for(int i=n;i>=1;--i) 57 { 58 sum[i]=res,res+=c[a[i].id]*num[a[i].id]; 59 sb[i]=res2,res2+=num[a[i].id]; 60 } 61 res=INF; 62 for(int i=1;i<=n;++i) 63 { 64 int kk=r[i];ll ttt=0; 65 if (r[i]!=r[i+1]) ttt=num[a[i].id]; 66 else{ 67 intJ = I; 68 the while (R & lt [J] == R & lt [J + . 1 ] && J <= n-) TTT + = NUM [A [J ++ ] .id]; 69 IF (J <= n-) TTT + = NUM [A [ J] .id]; 70 } 71 is LL-CNT = READ_A SB [KK] -ttt * 2 + . 1 ; 72 LL tmp = SUM [KK]; 73 is IF (CNT> 0 && R & lt [I] = R & lt [I-! . 1 ]) { 74 tmp + = Query (cnt, . 1 , Max, . 1 ); // delete the minimum cost cnt number 75 RES = min (RES, tmp); 76 } 77 IF (CNT <= 0 ! && R & lt [I] = R & lt [I- . 1 ]) RES = min (RES, tmp); 78 Update (C [A [I] .id], NUM [A [I] .id], . 1 , Max, . 1 ); // Add to this the cost of num. 79 } 80 the printf ( " % LLD \ n- " , RES); 81 } 82 return 0 ; 83 }
D.Number
Transmission: https://ac.nowcoder.com/acm/contest/887/D
The meaning of problems: a request output $ $ n-bit prime number multiples of $ p $ tree, there is no output T_T.
Data range: $ 1 <= n <= 10 ^ 6,2 <= p <= 10 ^ 6 $.
Analysis: direct construction. The output of a $ p $, the rest of the output 0.
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 int n,p;scanf("%d%d",&n,&p); 7 int tmp=0,x=p; 8 while (x) {x/=10;tmp++;} 9 if(n<tmp) printf("T_T\n"); 10 else 11 { 12 printf("%d",p); 13 for (int i=1;i<=n-tmp;i++) printf("0"); 14 printf("\n"); 15 } 16 return 0; 17 }
J.A+B problem
Transmission: https://ac.nowcoder.com/acm/contest/887/J
Meaning of the questions: $ number $ x $ reversed after the definition of $ f (x). Solving $ f (f (a) + f (b)) $.
Analysis: violence.
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 char a[50],b[50]; 5 int ans[50]; 6 int main(){ 7 int t;scanf("%d",&t); 8 while (t--){ 9 scanf("%s%s",a,b); 10 int lena=strlen(a),lenb=strlen(b); 11 ll f=1ll; ll aa=0,bb=0; 12 for (int i=0;i<lena;i++){ 13 aa=aa+f*(a[i]-'0'); 14 f*=10; 15 } 16 f=1ll; 17 for (int i=0;i<lenb;i++){ 18 bb=bb+f*(b[i]-'0'); 19 f*=10; 20 } 21 ll kk=aa+bb; int tot=0; 22 while (kk){ 23 ans[++tot]=kk%10; 24 kk/=10; 25 } 26 int i=1; 27 while (ans[i]==0) i++; 28 for (i;i<=tot;i++) printf("%d",ans[i]); 29 printf("\n"); 30 } 31 return 0; 32 }