2017-10-湖南套题2

 

 

处理0的前缀和，枚举第i位不变，[1,i-1]全变为0，[i+1,n]全变为1的最小代价

 1 #include <cstring>
 2 #include <cstdio>
 3 
 4 const int N(1e5+5);
 5 int sum[N],ans,tmp;
 6 char s[N];
 7 
 8 int Presist()
 9 {
10     freopen("reverse.in","r",stdin);
11     freopen("reverse.out","w",stdout);
12     scanf("%s",s); int n=strlen(s);
13     for(int i=1; i<=n; ++i)
14         sum[i]=sum[i-1]+(s[i-1]=='0');
15     ans=N;
16     for(int i=1; i<=n; ++i)
17     {
18         tmp=i-1-sum[i-1];
19         tmp+=sum[n]-sum[i];
20         ans=ans<tmp?ans:tmp;
21     }
22     printf("%d\n",ans);
23     return 0;
24 }
25 
26 int Aptal=Presist();
27 int main(int argc,char**argv){;}
AC

 

 

 

 

 

用hash的思想，出现字符相同的[1,n]中的每个数放到同一个集合里，组合数统计答案

 1 #include <algorithm>
 2 #include <cstdio>
 3 
 4 #define LL long long
 5 
 6 inline void read(int &x)
 7 {
 8     x=0; register char ch=getchar();
 9     for(; ch>'9'||ch<'0'; ) ch=getchar();
10     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
11 }
12 
13 const int N(1e7+5);
14 int p[10]={97,89,83,79,73,71,67,61,59,53};
15 bool vis[N][10];
16 LL cnt[N],ans,maxn;
17 int n;
18 
19 int Presist()
20 {
21 //    freopen("number.in","r",stdin);
22 //    freopen("number.out","w",stdout);
23     read(n);
24     for(int t,x,i=1; i<=n; ++i)
25     {
26         LL tmp=0;
27         for(x=i; x; x/=10)
28         {
29             t=x%10;
30             if(!vis[i][t]) tmp+=(t+101)*p[t];
31             vis[i][t]=1;
32         }
33         maxn=maxn>tmp?maxn:tmp;
34         cnt[tmp]++;
35     }
36     for(int i=101; i<=maxn; ++i)
37         if(cnt[i]>1) ans+=cnt[i]*(cnt[i]-1)>>1;
38     printf("%I64d\n",ans);
39     return 0;
40 }
41 
42 int Aptal=Presist();
43 int main(int argc,char**argv){;}
AC

 

 1 #include<cstdio>
 2 #define maxn 2000
 3 using namespace std;
 4 int n,v[maxn];
 5 long long ans;
 6 int main()
 7 {
 8     freopen("number.in","r",stdin);
 9     freopen("number.out","w",stdout);
10     int a,b,x;
11     scanf("%d",&n);
12     for(int i=1;i<=n;i++)
13     {
14         a=i;
15         x=0;
16         while(a)
17         {
18             b=a%10;
19             x|=1<<b;
20             a=a/10;
21         }
22         ans+=v[x];
23         v[x]++;
24     }
25     printf("%I64d\n",ans);
26     return 0;
27 }
std

 

 

 

 

 

贪心：满足相邻两个数的差>=k，偶数项尽量大，奇数项尽量小

 1 #include <cstring>
 2 #include <cstdio>
 3 
 4 inline void read(int &x)
 5 {
 6     x=0; register char ch=getchar();
 7     for(; ch>'9'||ch<'0'; ) ch=getchar();
 8     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
 9 }
10 const int N(2e6+5);
11 int n,k,a[N];
12 
13 int Presist()
14 {
15 //    freopen("wave.in","r",stdin);
16 //    freopen("wave.out","w",stdout);
17     read(n),read(k);
18     for(int i=1; i<=n; ++i) read(a[i]);
19     int op=0,pre=a[1],ans=1;
20     for(int i=2; i<=n; ++i)
21     {
22         if(op)
23             if(pre-a[i]>=k)
24                 op=0,pre=a[i],ans++,printf("%d ",pre);
25             else pre=pre>a[i]?pre:a[i];
26         else if(a[i]-pre>=k)
27                 op=1,pre=a[i],ans++,printf("%d ",pre);
28              else pre=pre<a[i]?pre:a[i];
29     }
30     printf("%d\n",ans);
31     return 0;
32 }
33 
34 int Aptal=Presist();
35 int main(int argc,char**argv){;}
AC

 

posted @ 2017-10-17 15:51  Aptal丶  阅读(109)  评论(0)  编辑  收藏  举报