Codeforces Round #332 (Div. 2)
颓得不行，于是没打，第二天早上爬起来码代码
- A
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 int main(){ 5 int a,b,c; 6 scanf("%d%d%d",&a,&b,&c); 7 int q=a+b+c; 8 int w=(min(a,b)+c)*2; 9 int e=(a+b)*2; 10 printf("%d\n",min(q,min(w,e))); 11 return 0; 12 }
- B
1 #include<stdio.h> 2 #define maxn 100005 3 int n,m,f[maxn],b[maxn],Map[maxn],tong[maxn],a[maxn]; 4 int main(){ 5 scanf("%d%d",&n,&m); 6 for(int i=1;i<=n;i++){ 7 scanf("%d",&f[i]); 8 Map[f[i]]=i; 9 tong[f[i]]++; 10 } 11 bool amb=false,imp=false; 12 for(int i=1;i<=m;i++){ 13 scanf("%d",&b[i]); 14 if(!tong[b[i]])imp=true; 15 if(tong[b[i]]>1)amb=true; 16 a[i]=Map[b[i]]; 17 } 18 if(!imp&&!amb){ 19 printf("Possible\n"); 20 for(int i=1;i<=m;i++) 21 printf("%d ",a[i]); 22 printf("\n"); 23 } 24 else if(imp)printf("Impossible\n"); 25 else printf("Ambiguity\n"); 26 return 0; 27 }
- C
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 5 #define maxn 100005 6 int n,yooo[maxn],Maxyooo[maxn],Minyooo[maxn]; 7 8 int main(){ 9 //freopen("1.in","r",stdin); 10 scanf("%d",&n); 11 for(int i=1;i<=n;i++){ 12 scanf("%d",&yooo[i]); 13 Maxyooo[i]=max(Maxyooo[i-1],yooo[i]); 14 } 15 Minyooo[n-1]=yooo[n]; 16 for(int i=n-2;i>=1;i--) 17 Minyooo[i]=min(Minyooo[i+1],yooo[i+1]); 18 int ans=1; 19 for(int i=1;i<=n;i++) 20 if(Maxyooo[i]<=Minyooo[i])ans++; 21 printf("%d\n",ans); 22 return 0; 23 }
前三题做完发现，打cf还是不要想太多，因为每次只有前【两个】想法是对的。。。
- D
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 5 typedef long long i64; 6 #define fir first 7 #define sec second 8 #define mp make_pair 9 #define maxn 1000005 10 i64 x,squ,res,cnt; 11 pair<i64,i64>ans[maxn]; 12 int main(){ 13 //freopen("1.in","r",stdin); 14 scanf("%I64d",&x); 15 for(i64 n=1;squ=n*(n+1)*(2*n+1)/6,squ<=x;n++){ 16 i64 res=n*(n+1)/2; 17 if((x-squ)%res==0)ans[++cnt]=mp(n,n+(x-squ)/res); 18 } 19 bool flag=false; 20 if(ans[cnt].fir==ans[cnt].sec)flag=true; 21 printf("%I64d\n",flag?cnt*2-1:cnt*2); 22 for(i64 i=1;i<=cnt;i++) 23 printf("%I64d %I64d\n",ans[i].fir,ans[i].sec); 24 for(i64 i=cnt;i>=1;i--){ 25 if(flag&&i==cnt)continue; 26 else printf("%I64d %I64d\n",ans[i].sec,ans[i].fir); 27 } 28 return 0; 29 }
略美，然而并不知道1^2+2^2+...+n^2=n*(n+1)*(2*n+1)/6，在这里跪谢wh大爷
- E South Africa
