Codeforces Round #142 (Div. 2)
2012-10-02 13:52
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A | Dragons standard input/output 2 s, 256 MB | x1446 |
看到此题之后再看到数据比较小,就直接暴力O(n^2).
每次寻找能杀的龙之中能得到能力提升最大的一只即可。
#include <algorithm> #include <iostream> #include <fstream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <stack> #include <queue> #include <set> #include <map> #define inf (1<<31)-1 #define PI 3.141592653 #define mod 1000000007 #define N 1000005 #define M 1000005 #define pb(a) push_back(a); #define empty em #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; int a ,b ,h ; int s,n; int main() { while(scanf("%d%d",&s,&n)!=EOF) { for(int i=1;i<=n;i++)scanf("%d%d",a+i,b+i); bool flag=1; mem(h,0); for(int i=1;i<=n;i++) { int num,maxx=-1,ff=0,t1,t2; for(int j=1;j<=n;j++) { if(s>a[j]&&maxx<b[j]&&h[j]==0) { t1=b[j]; t2=j; ff=1; } } if(ff==0) { flag=0; break; } else { s+=t1; h[t2]=1; } } puts(flag?"YES":"NO"); } return 0; }
B | T-primes standard input/output 2 s, 256 MB | x1105 |
假设数x是一个满足的数,它的因子为a,b,c,此时可以假定a=1,b=x,那么c为多少呢,设d=x/c,如果x!=c^2,那么d一定为一个新的因子,所以c==d,所以我们可以知道x一定是一个素数的平方数,到此题目就很明确了!
#include <algorithm> #include <iostream> #include <fstream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <stack> #include <queue> #include <set> #include <map> #define inf (1<<31)-1 #define PI 3.141592653 #define mod 1000000007 #define N 4000005 #define M 10000005 #define pb(a) push_back(a); #define empty em #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const ll MAXN=4000005; ll prime[MAXN]; int cnt; void getprime() { memset(prime,0,sizeof(prime)); for(int i=2;i<=MAXN;i++) { if(prime[i])continue; for(int j=2;j*i<=MAXN;j++)prime[j*i]=1; } cnt=0; for(int i=2;i<=MAXN;i++) if(!prime[i])prime[++cnt]=i; } int main() { getprime(); int n; while(scanf("%d",&n)!=EOF) { map<ll,int> ma; for(ll i=1;prime[i]<=1000000;i++) ma[prime[i]*prime[i]]=1; for(int i=1;i<=n;i++) { ll x; scanf("%I64d",&x); puts(ma[x]?"YES":"NO"); } } return 0; }
C | Shifts standard input/output 2 s, 256 MB | x509 |
设置数组le[i][j](代表第i行第j列的元素在第i行能从左向右平移到j位置的最小操作数),ri[i][j]同理,re[i][j]则为总的最小的,由于可以从尾部向首部循环移动,所以可以加倍数组长度,复制前半部分到后半部分,这样就可以实现循环了,然后加上没一列的rep[i][j],取最小即可,中间记录一下一横行全为0的情况,这样就可以直接输出零。
#include <algorithm> #include <iostream> #include <fstream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <stack> #include <queue> #include <set> #include <map> #define inf (1<<31)-1 #define PI 3.141592653 #define mod 1000000007 #define N 1000005 #define M 1000005 #define pb(a) push_back(a); #define empty em #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; char ma[105][20005]; int le[105][20005]; int ri[105][20005]; int re[105][20005]; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { mem(le,0);mem(ri,0);mem(re,0); bool flag=1; int ans=(1<<31)-1; for(int i=1;i<=n;i++)scanf("%s",ma[i]+1); for(int i=1;i<=n;i++) for(int j=m+1;j<=2*m;j++) ma[i][j]=ma[i][j-m]; for(int i=1;i<=n;i++) { for(int j=1;j<=2*m;j++) { if(ma[i][j]=='1')le[i][j]=j; else le[i][j]=le[i][j-1]; } for(int j=2*m;j>=1;j--) { if(ma[i][j]=='1')ri[i][j]=j; else ri[i][j]=ri[i][j+1]; } int total=0; for(int j=1;j<=2*m;j++) { if(le[i][j]!=0&&ri[i][j]!=0)re[i][j]=min(j-le[i][j],ri[i][j]-j); else if(le[i][j]==0&&ri[i][j]!=0)re[i][j]=ri[i][j]-j; else if(le[i][j]!=0&&ri[i][j]==0)re[i][j]=j-le[i][j]; else total++; } if(total==2*m){flag=0;goto ll;} } ll : for(int j=1;j<=2*m;j++) { int tmp=0; for(int i=1;i<=n;i++) tmp+=re[i][j]; ans=min(ans,tmp); } if(flag)printf("%d\n",ans); else printf("-1\n"); } return 0; }
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