2015 Multi-University Training Contest 9

1001 Expression

式子不好推啊。见官方题解。

式子知道就方便了。处理好组合数和阶乘。

按区间长度从小到大递推完就好。

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 using namespace std;
 5 # define maxn 105
 6 typedef long long LL;
 7 const LL mod=1e9+7;
 8 LL fac[maxn]={1},C[maxn][maxn];
 9 LL a[maxn],dp[maxn][maxn];
10 char op[maxn];
11
12 int main(void)
13 {
14     for(int i=1;i<maxn;i++) fac[i]=(fac[i-1]*(LL)i)%mod;
15     for(int i=0;i<maxn;i++) C[i][0]=(LL)1;
16     for(int i=1;i<maxn;i++)
17         for(int j=1;j<=i;j++)
18             C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
19     int n;
20     while(~scanf("%d",&n))
21     {
22         memset(dp,0,sizeof(dp));
23         for(int i=1;i<=n;i++) scanf("%I64d",a+i);
24         scanf("%s",op+1);
25         for(int i=1;i<=n;i++) dp[i][i]=a[i];
26         for(int len=2;len<=n;len++)
27         {
28             for(int s=1;s+len-1<=n;s++)
29             {
30                 for(int m=s;m<s+len-1;m++)
31                 {
32                     LL tem;
33                     if(op[m]=='*') tem=dp[s][m]*dp[m+1][s+len-1];
34                     else if(op[m]=='+') tem=dp[s][m]*fac[s+len-m-2]+dp[m+1][s+len-1]*fac[m-s];
35                     else tem=dp[s][m]*fac[s+len-m-2]-dp[m+1][s+len-1]*fac[m-s];
36                     dp[s][s+len-1]+=tem%mod*C[len-2][m-s]%mod;
37                 }
38                 dp[s][s+len-1]=(dp[s][s+len-1]%mod+mod)%mod;
39             }
40         }
41         printf("%I64d\n",dp[1][n]);
42     }
43     return 0;
44 }

Aguin

1002 Hack it!

1003 GCD Tree

1004 Too Simple

果然还是图样。置换顺序乘反了。

比赛wa到底。拿了数据才知道错在哪。 QAQ。

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 using namespace std;
 5 typedef long long LL;
 6 const LL mod=1e9+7;
 7 int a[101][101],vis[101];
 8 LL fac[101]={1};
 9
10 LL qpow(LL a,int b)
11 {
12     LL d=(LL)1,t=a;
13     while(b)
14     {
15         if(b%2) d=(d*t)%mod;
16         b/=2;
17         t=(t*t)%mod;
18     }
19     return d;
20 }
21
22 int main(void)
23 {
24     for(int i=1;i<=100;i++) fac[i]=(fac[i-1]*LL(i))%mod;
25     int n,m;
26     while(~scanf("%d%d",&n,&m))
27     {
28         int ok=1,cnt=0;
29         for(int i=1;i<=m;i++)
30         {
31             scanf("%d",&a[i][1]);
32             if(a[i][1]==-1) cnt++;
33             else
34             {
35                 for(int j=2;j<=n;j++) scanf("%d",&a[i][j]);
36                 memset(vis,0,sizeof(vis));
37                 for(int j=1;j<=n;j++) vis[a[i][j]]++;
38                 for(int j=1;j<=n;j++) if(vis[j]!=1) ok=0;
39             }
40         }
41         if(!cnt) for(int i=1;i<=n;i++)
42         {
43             int pos=i;
44             for(int j=m;j>0;j--) pos=a[j][pos];
45             if(pos!=i) ok=0;
46         }
47         if(!ok) puts("0");
48         else
49         {
50             LL ans=1;
51             if(cnt) ans=qpow(fac[n],cnt-1);
52             printf("%I64d\n",ans);
53         }
54     }
55     return 0;
56 }

Aguin

1005 Arithmetic Sequence

以每个点为i。处理一下向前向后能延伸的最大距离。

d1≠d2时。答案为sigma(pre[i]*suf[i])。

d1=d2时。pre[i]相当于以i为右端点的区间数。答案为sigma(pre[i])。

 1 # include <iostream>
 2 # include <cstdio>
 3 using namespace std;
 4 typedef long long LL;
 5 const int maxn=100100;
 6 int a[maxn],pre[maxn],suf[maxn];
 7
 8 int main(void)
 9 {
10     int n,d1,d2;
11     while(~scanf("%d%d%d",&n,&d1,&d2))
12     {
13         for(int i=1;i<=n;i++) scanf("%d",a+i);
14         pre[1]=1; suf[n]=1;
15         for(int i=2;i<=n;i++)
16         {
17             if(a[i]==a[i-1]+d1) pre[i]=pre[i-1]+1;
18             else pre[i]=1;
19         }
20         for(int i=n-1;i>0;i--)
21         {
22             if(a[i]==a[i+1]-d2) suf[i]=suf[i+1]+1;
23             else suf[i]=1;
24         }
25         LL ans=0;
26         if(d1==d2) for(int i=1;i<=n;i++) ans+=(LL)pre[i];
27         else for(int i=1;i<=n;i++) ans+=(LL)pre[i]*(LL)suf[i];
28         printf("%I64d\n",ans);
29     }
30     return 0;
31 }

Aguin

1006 Persistent Link/cut Tree

1007 Travelling Salesman Problem

n或者m为奇数时必然能走完。

全偶数时选择行标加列标为奇数的格子中最小的格子不走。其余都能走。

（因为是从偶数格子出发到偶数格子终止的。所以不存在只选一个行标加列标为偶数的格子不走。其他格子都走的情况。）

具体做法是。设选中格子坐标为(i,j)。如果i为奇。可以先走到（i,j-1）。再绕过选中格子。绕到（i,j+2）。就转变为一边为奇数的情况了。

同理i为偶数先走到(i-1,j)。绕到(i+2,j)。按奇数边的走完。

要注意选中格子i==n或者j==m的情况要特判一下。

【代码丑- -】

  1 # include <iostream>
  2 # include <cstdio>
  3 using namespace std;
  4 # define INF 2147483647
  5 int n,m,map[101][101];
  6
  7 void Print_char(int len,char c)
  8 {
  9     for(int i=0;i<len;i++) putchar(c);
 10 }
 11
 12 void ans_print(int x,int y)
 13 {
 14     if((n-x)%2==0)
 15     {
 16         int len=m-y;
 17         for(int i=x;i<n;i+=2)
 18         {
 19             Print_char(len,'R');
 20             putchar('D');
 21             Print_char(len,'L');
 22             putchar('D');
 23         }
 24         Print_char(len,'R');
 25     }
 26     else
 27     {
 28         int len=n-x;
 29         for(int i=y;i<m;i+=2)
 30         {
 31             Print_char(len,'D');
 32             putchar('R');
 33             Print_char(len,'U');
 34             putchar('R');
 35         }
 36         Print_char(len,'D');
 37     }
 38     printf("\n");
 39     return;
 40 }
 41
 42 int main(void)
 43 {
 44     while(~scanf("%d%d",&n,&m))
 45     {
 46         int sum=0;
 47         for(int i=1;i<=n;i++)
 48         {
 49             for(int j=1;j<=m;j++)
 50             {
 51                 scanf("%d",&map[i][j]);
 52                 sum+=map[i][j];
 53             }
 54         }
 55         if(n%2||m%2)
 56         {
 57             printf("%d\n",sum);
 58             ans_print(1,1);
 59             continue;
 60         }
 61         int Min=INF,x,y;
 62         for(int i=1;i<=n;i++)
 63             for(int j=1;j<=m;j++)
 64                 if((i+j)%2==1&&map[i][j]<Min)
 65                     {Min=map[i][j];x=i;y=j;}
 66         printf("%d\n",sum-Min);
 67         int xx=1,yy=1;
 68         if(x%2==0)
 69         {
 70             while(xx<x-1)
 71             {
 72                 int len=m-yy;
 73                 Print_char(len,'R');
 74                 putchar('D');
 75                 Print_char(len,'L');
 76                 putchar('D');
 77                 xx+=2;
 78             }
 79             while(yy<y)
 80             {
 81                 int len=n-xx;
 82                 Print_char(len,'D');
 83                 putchar('R');
 84                 Print_char(len,'U');
 85                 putchar('R');
 86                 yy+=2;
 87             }
 88             if(x<n)
 89             {
 90                 Print_char(m-yy,'R');
 91                 putchar('D');
 92                 for(int i=m;i>yy;i--)
 93                 {
 94                     if((i-yy)%2) printf("DL");
 95                     else printf("UL");
 96                 }
 97                 putchar('D');
 98                 ans_print(x+2,y);
 99             }
100             else
101             {
102                 putchar('R');
103                 ans_print(x-1,y+1);
104             }
105         }
106         else
107         {
108             while(xx<x)
109             {
110                 int len=m-yy;
111                 Print_char(len,'R');
112                 putchar('D');
113                 Print_char(len,'L');
114                 putchar('D');
115                 xx+=2;
116             }
117             while(yy<y-1)
118             {
119                 int len=n-xx;
120                 Print_char(len,'D');
121                 putchar('R');
122                 Print_char(len,'U');
123                 putchar('R');
124                 yy+=2;
125             }
126             if(y<m)
127             {
128                 Print_char(n-x,'D');
129                 putchar('R');
130                 for(int i=n;i>xx;i--)
131                 {
132                     if((i-xx)%2) printf("RU");
133                     else printf("LU");
134                 }
135                 putchar('R');
136                 ans_print(x,y+2);
137             }
138             else
139             {
140                 putchar('D');
141                 ans_print(x+1,y-1);
142             }
143         }
144     }
145     return 0;
146 }

Aguin

1008 Goldbach's Conjecture

1009 Random Inversion Machine

1010 Sometimes Naive

转载于:https://www.cnblogs.com/Aguin/p/4740083.html

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