code+11月月赛
T1 SB题
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<ctime> 5 #include<iostream> 6 #include<algorithm> 7 #include<queue> 8 #include<set> 9 #define inf (0x7fffffff) 10 #define l(a) ((a)<<1) 11 #define r(a) ((a)<<1|1) 12 #define b(a) (1<<(a)) 13 #define rep(i,a,b) for(int i=a;i<=(b);i++) 14 #define clr(a) memset(a,0,sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 using namespace std; 18 ll a,b,c; 19 ll gcd(ll x,ll y){ 20 return y?gcd(y,x%y):x; 21 } 22 int main(){ 23 scanf("%lld%lld%lld",&a,&b,&c); 24 ll t=a/gcd(a,b)*b; 25 cout<<c/gcd(t,c)*t<<endl; 26 return 0; 27 }
T2 Sort贪心 注意边界和特殊情况
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<ctime> 5 #include<iostream> 6 #include<algorithm> 7 #include<queue> 8 #include<set> 9 #define inf (0x7fffffff) 10 #define l(a) ((a)<<1) 11 #define r(a) ((a)<<1|1) 12 #define b(a) (1<<(a)) 13 #define rep(i,a,b) for(int i=a;i<=(b);i++) 14 #define clr(a) memset(a,0,sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 using namespace std; 18 const int maxn=200009; 19 ll n,S,L,sum[maxn],H[maxn]; 20 struct tree{ 21 ll h,a,m; 22 inline bool operator<(const tree T)const{ 23 return m<T.m||(m==T.m&&a<T.a); 24 } 25 }t[maxn]; 26 int main(){ 27 scanf("%lld%lld%lld",&n,&S,&L); 28 rep(i,1,n) scanf("%lld",&t[i].h); 29 rep(i,1,n) scanf("%lld",&t[i].a); 30 rep(i,1,n) t[i].m=L<=t[i].h?0:(L-t[i].h)/t[i].a+ll((L-t[i].h)%t[i].a!=0); 31 sort(t+1,t+n+1); 32 rep(i,1,n){ 33 sum[i]=sum[i-1]+t[i].a; 34 H[i]=H[i-1]+t[i].h; 35 } 36 bool flag=0; 37 rep(i,1,n){ 38 ll tmp=t[i].m*sum[i]+H[i]; 39 if(tmp>=S){ 40 flag=1; 41 if(i==1) cout<<max(t[1].m,S<=H[1]?0:(S-H[1])/t[1].a+ll((S-H[1])%t[1].a!=0))<<endl; 42 else cout<<min(t[i].m,(S-H[i-1])/sum[i-1]+ll((S-H[i-1])%sum[i-1]!=0))<<endl; 43 break; 44 } 45 } 46 if(!flag) cout<<(S<=H[n]?0:(S-H[n])/sum[n]+ll((S-H[n])%sum[n]!=0))<<endl; 47 return 0; 48 }
T3 DP 我tm把m打成n 一直wa
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<ctime> 5 #include<iostream> 6 #include<algorithm> 7 #include<queue> 8 #include<set> 9 #define inf (0x3f3f3f3f) 10 #define l(a) ((a)<<1) 11 #define r(a) ((a)<<1|1) 12 #define b(a) (1<<(a)) 13 #define rep(i,a,b) for(int i=a;i<=(b);i++) 14 #define clr(a) memset(a,0,sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 using namespace std; 18 int readint(){ 19 int t=0,f=1;char c=getchar(); 20 while(!isdigit(c)){ 21 if(c=='-') f=-1; 22 c=getchar(); 23 } 24 while(isdigit(c)){ 25 t=(t<<3)+(t<<1)+c-'0'; 26 c=getchar(); 27 } 28 return t*f; 29 } 30 const int maxn=6009; 31 int n,m,_A,_B,a[maxn],b[maxn],c[5][5],d[maxn][maxn][3]; 32 char A[maxn],B[maxn]; 33 int f(char c){ 34 if(c=='A') return 1; 35 if(c=='T') return 2; 36 if(c=='G') return 3; 37 return 4; 38 } 39 int main(){ 40 scanf("%s%s",A,B);n=strlen(A);m=strlen(B); 41 rep(i,1,n) a[i]=f(A[i-1]); 42 rep(i,1,m) b[i]=f(B[i-1]); 43 rep(i,1,4) 44 rep(j,1,4) c[i][j]=readint(); 45 _A=readint();_B=readint();_A=-_A;_B=-_B; 46 rep(i,0,n) 47 rep(j,0,m) 48 rep(k,0,2) d[i][j][k]=-inf; 49 d[0][0][0]=0; 50 rep(i,0,n){ 51 rep(j,0,m){ 52 rep(k,0,2) if(i>0&&j>0) d[i][j][0]=max(d[i][j][0],d[i-1][j-1][k]+c[a[i]][b[j]]); 53 if(j){ 54 d[i][j][1]=max(d[i][j][1],d[i][j-1][0]+_A); 55 d[i][j][1]=max(d[i][j][1],d[i][j-1][1]+_B); 56 d[i][j][1]=max(d[i][j][1],d[i][j-1][2]+_A); 57 } 58 if(i){ 59 d[i][j][2]=max(d[i][j][2],d[i-1][j][0]+_A); 60 d[i][j][2]=max(d[i][j][2],d[i-1][j][2]+_B); 61 d[i][j][2]=max(d[i][j][2],d[i-1][j][1]+_A); 62 } 63 } 64 } 65 int ans=-inf; 66 rep(i,0,2) ans=max(ans,d[n][m][i]); 67 printf("%d\n",ans); 68 return 0; 69 }
T4 并不难的DP...
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<ctime> 5 #include<iostream> 6 #include<algorithm> 7 #include<queue> 8 #include<set> 9 #define inf (0xffffffffll) 10 #define l(a) ((a)<<1) 11 #define r(a) ((a)<<1|1) 12 #define b(a) (1<<(a)) 13 #define rep(i,a,b) for(int i=a;i<=(b);i++) 14 #define clr(a) memset(a,0,sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 using namespace std; 18 ll readll(){ 19 ll t=0,f=1;char c=getchar(); 20 while(!isdigit(c)){ 21 if(c=='-') f=-1; 22 c=getchar(); 23 } 24 while(isdigit(c)){ 25 t=(t<<3ll)+(t<<1ll)+ll(c-'0'); 26 c=getchar(); 27 } 28 return ll(t*f); 29 } 30 const int maxm=1000009; 31 ll n,m,d[maxm],a[maxm],q[maxm]; 32 struct node{ 33 ll p,w; 34 inline bool operator<(const node A)const{ 35 return p<A.p; 36 } 37 inline void in(){ 38 p=readll();w=readll(); 39 } 40 }x[maxm]; 41 int main(){ 42 //freopen("#input.txt","r",stdin); 43 //freopen("#output.txt","w",stdout); 44 n=readll();m=readll(); 45 rep(i,1,maxm-1) q[i]=-1; 46 rep(i,1,m) x[i].in();sort(x+1,x+m+1); 47 n=0;rep(i,1,m){ 48 if(x[i].p==x[i-1].p+1){ 49 q[++n]=x[i].w; 50 }else if(x[i].p==x[i-1].p+2){ 51 n+=2;q[n]=x[i].w; 52 }else{ 53 n+=3;q[n]=x[i].w; 54 } 55 } 56 d[0]=0; 57 rep(i,1,n) if(q[i]!=-1){ 58 a[i]=a[i-1]^q[i]; 59 d[i]=d[i-1]+a[i]; 60 }else if(q[i+1]!=-1){ 61 int cnt=0;ll t1=inf,s1[32],t2=0,s2[32]; 62 clr(s1);clr(s2); 63 for(int j=i+1;q[j]!=-1;j++,cnt++){ 64 t1^=q[j]; 65 rep(l,0,31) if((1ll<<l)&t1) s1[l]++; 66 } 67 for(int j=i+1;q[j]!=-1;j++){ 68 t2^=q[j]; 69 rep(l,0,31) if((1ll<<l)&t2) s2[l]++; 70 } 71 i+=cnt; 72 d[i]=d[i-cnt-1]; 73 rep(j,0,31){ 74 t1=s1[j]+((a[i-cnt]^(1<<j))>>j);t2=s2[j]+((a[i-cnt]&(1<<j))>>j); 75 if(t1>t2){ 76 d[i]+=t2*(1ll<<j);a[i]+=t2&(1ll<<j); 77 }else{ 78 d[i]+=t1*(1ll<<j);a[i]+=t1&(1ll<<j); 79 } 80 } 81 }else d[i]=d[i-1],a[i]=0; 82 printf("%lld\n",d[n]); 83 //fclose(stdin); 84 //fclose(stdout); 85 return 0; 86 }
