题解

99 条题解

  • 2
    @ 2017-12-11 19:11:09

    环形dp,要义在于破坏成链。两倍链式展开再dp,最后扫一遍答案。
    b[i][j][k]表示区间[i,j]分成k段的最大值,s[i][j][k]表示区间[i,j]分成k段的最小值。
    注意负数取模。本题可用前缀和优化。
    初始化时注意每个区间仅分为一段时为1。

    
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    int n,m,a[101],b[101][101][10],s[101][101][10];
    int mo(int x)
    {
        return (x%10+10)%10;
    }
    int main()
    {
        int i,j,l,r,maxn=0,minn=0x7777777;
        scanf("%d%d",&n,&m);
        for (i=1;i<=n;++i)
        {
            scanf("%d",&a[i]);
            a[i+n]=a[i];
        }
        for (i=1;i<=n*2;++i)
            a[i]+=a[i-1];
        memset(s,233333,sizeof(s));
        for (l=1;l<=n*2;++l)
            for (r=l;r<=2*n;++r)
                b[l][r][1]=s[l][r][1]=mo(a[r]-a[l-1]);
        for (i=2;i<=m;++i)
            for (l=1;l<=n*2;++l)
                for (r=l+i-1;r<=n*2;++r)
                    for (j=l+i-2;j<r;++j)
                    {
                        b[l][r][i]=max(b[l][r][i],b[l][j][i-1]*mo(a[r]-a[j]));
                        s[l][r][i]=min(s[l][r][i],s[l][j][i-1]*mo(a[r]-a[j]));
                    }
        for (i=1;i<=n;++i)
        {
            maxn=max(b[i][i+n-1][m],maxn);
            minn=min(s[i][i+n-1][m],minn);
        }
        printf("%d\n%d",minn,maxn);
        return 0;
    }
    
    
    
  • 1
    @ 2017-10-01 19:06:13

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    int jyx[100],a[100],yx[10];
    int n,m;
    long long sum=0x7fffffff,maxsum=-1;
    int print(long long ksum)
    {
    sum=min(sum,ksum);
    maxsum=max(maxsum,ksum);
    }
    int locin(int k,int b,long long nsum)
    {
    for(int i=b;i<=n+k-m;++i)
    {
    long long ksum=nsum;
    yx[k]=i;
    if(k==1)locin(k+1,i+1,nsum);
    else
    {
    long long h1=jyx[i]-jyx[yx[k-1]];
    while(h1<0)h1+=10;
    ksum*=(h1%10);
    if(k==m)
    {
    long long h=jyx[yx[1]]+jyx[n]-jyx[i];
    while(h<0)h+=10;
    ksum*=(h%10);
    print(ksum);
    }
    else locin(k+1,i+1,ksum);
    }
    }
    }
    int main()
    {
    cin>>n>>m;
    for(int i=1;i<=n;++i)cin>>a[i],jyx[i]=jyx[i-1]+a[i];
    if(m==1)
    {
    while(jyx[n]<0)jyx[n]+=10;
    cout<<jyx[n]%10<<endl<<jyx[n]%10;
    return 0;
    }
    locin(1,1,1);
    cout<<sum<<endl<<maxsum;
    return 0;
    }

  • 0
    @ 2019-03-12 20:13:26

    细节问题上wa了N次,哭了
    ```cpp
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    using namespace std;
    int N,M;
    int a[110],d[10][110][110],vis[10][110][110],smax,smin; //d[m][i][j] 区间[i,j]分为m段的最大值
    int dpmax(int m,int i,int j){ //闭区间
    int &ans = d[m][i][j];
    if(vis[m][i][j]) return ans;
    vis[m][i][j] = 1;
    if(m==1){
    ans = 0;
    for(int ii=i;ii<=j;ii++) ans += a[ii];
    ans = (ans%10+10)%10;
    return ans;
    }
    for(int k=i;k<j;k++){
    if(m-1>0&&m-1<=j-k) ans = max(ans,dpmax(1,i,k)*dpmax(m-1,k+1,j));
    else ans = max(ans,dpmax(1,i,k));
    }
    return ans;
    }
    int dpmin(int m,int i,int j){ //闭区间
    int &ans = d[m][i][j];
    if(vis[m][i][j]) return ans;
    vis[m][i][j] = 1;
    if(m==1){
    ans = 0;
    for(int ii=i;ii<=j;ii++) ans += a[ii];
    ans = (ans%10+10)%10;
    return ans;
    }
    for(int k=i;k<j;k++){
    if(m-1>0&&m-1<=j-k) ans = min(ans,dpmin(1,i,k)*dpmin(m-1,k+1,j));
    else ans = min(ans,dpmin(1,i,k));
    }
    return ans;
    }
    int main(){
    ios::sync_with_stdio(false);
    cin>>N>>M;
    for(int i=0;i<N;i++){
    cin>>a[i];
    a[i+N] = a[i];
    }
    for(int i=0;i<N;i++) smax = max(smax,dpmax(M,i,i+N-1));
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=M;i++){
    for(int j=0;j<N*2;j++){
    for(int k=0;k<N*2;k++){
    d[i][j][k]= 0x3fffffff;
    }
    }
    }
    smin = 0x3fffffff;
    for(int i=0;i<N;i++) smin = min(smin,dpmin(M,i,i+N-1));
    cout<<smin<<'\n'<<smax;
    return 0;
    }

  • 0
    @ 2016-09-18 21:21:03

    搞了个DP
    评测结果
    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 2516 KiB, score = 20

    测试数据 #1: Accepted, time = 15 ms, mem = 2516 KiB, score = 20

    测试数据 #2: Accepted, time = 15 ms, mem = 2524 KiB, score = 20

    测试数据 #3: Accepted, time = 31 ms, mem = 2520 KiB, score = 20

    测试数据 #4: Accepted, time = 31 ms, mem = 2520 KiB, score = 20

    Accepted, time = 92 ms, mem = 2524 KiB, score = 100
    欢迎各位大神dalao到小弟博客看看,您的到来让本寒舍蓬荜生辉
    moe.cn.com
    代码
    #include<cstdio>
    #include<iostream>
    using namespace std;
    int n, m;
    int a[500];
    int sum[500];
    int dp[500][500];
    int dp2[500][500];
    int chuli(int a){
    a %= 10;
    if (a < 0) return (10 + a) % 10;
    return a % 10;
    }
    int main(){
    int ans = 0;
    int ans2 = 0x3f3f3f3f;
    cin >> n >> m;
    for (int i = 1; i <= n; i++){
    cin >> a[i];
    a[i + n] = a[i];
    }
    sum[0] = 0;
    for (int i = 1; i <= 2 * n; i++) sum[i] += sum[i - 1] + a[i];
    for (int s = 0; s < n; s++){//第几位开始
    for (int i = 0; i < 500; i++)
    for (int j = 0; j < 500; j++) {
    if (i == 0 || j == 0) dp[i][j] = 1, dp2[i][j] = 1;
    else dp[i][j] = -0x3f3f3f3f, dp2[i][j] = 0x3f3f3f3f;
    }
    for (int i = 1; i <= n; i++)//长度
    for (int j = 1; j <= min(i, m); j++)//断开几段
    {
    if (j == 1)//只有一段就不用枚举了
    {
    dp[i][j] = dp2[i][j] = chuli(sum[s + i] - sum[s]);
    continue;
    }
    for (int k = 1; k <= i - j + 1; k++){//长度枚举
    dp[i][j] = max(dp[i][j], dp[i - k][j - 1] * chuli(sum[s + i] - sum[s + i - k]));
    dp2[i][j] = min(dp2[i][j], dp2[i - k][j - 1] * chuli(sum[s + i] - sum[s + i - k]));
    }
    }
    ans = max(ans, dp[n][m]);
    ans2 = min(ans2, dp2[n][m]);

    }

    cout << ans2 << endl << ans << endl;
    }

  • 0
    @ 2016-08-31 18:05:07
    评测结果
    编译成功
    
    测试数据 #0: Accepted, time = 0 ms, mem = 1380 KiB, score = 20
    测试数据 #1: Accepted, time = 0 ms, mem = 1376 KiB, score = 20
    测试数据 #2: Accepted, time = 0 ms, mem = 1376 KiB, score = 20
    测试数据 #3: Accepted, time = 0 ms, mem = 1380 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 1380 KiB, score = 20
    Accepted, time = 0 ms, mem = 1380 KiB, score = 100
    代码
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    using namespace std;
    //ifstream cin("game.in",ios :: in);
    //ofstream cout("game.out",ios :: out);
    struct DP {
      int Max,Min;
    }dp[51*2][51*2][10];
    int n,M,a[51*2];
    inline int DfsMax(int L,int R,int m) {
      if (dp[L][R][m].Max != -9999999) return dp[L][R][m].Max;
      if (m == 1) {
        int sum = 0;
        for (int i = L;i < R;i++) sum += a[i];
        dp[L][R][1].Max = sum%10;
        return dp[L][R][1].Max;
      }
      int ans = -9999999;
      for (int i = L+1;i < R;i++) ans = max(ans,DfsMax(L,i,1)*DfsMax(i,R,m-1));
      dp[L][R][m].Max = ans;
      return dp[L][R][m].Max;
    }
    inline int DfsMin(int L,int R,int m) {
      if (dp[L][R][m].Min != 9999999) return dp[L][R][m].Min;
      if (m == 1) {
        int sum = 0;
        for (int i = L;i < R;i++) sum += a[i];
        dp[L][R][1].Min = sum%10;
        return dp[L][R][1].Min;
      }
      int ans = 9999999;
      for (int i = L+1;i < R;i++) ans = min(ans,DfsMin(L,i,1)*DfsMin(i,R,m-1));
      dp[L][R][m].Min = ans;
      return dp[L][R][m].Min;
    }
    int main() {
      ios :: sync_with_stdio(false);
      cin >> n >> M;
      for (int i = 1;i <= n*2;i++)
        for (int j = 1;j <= n*2;j++)
          for (int k = 1;k <= 9;k++) {
            dp[i][j][k].Max = -9999999;
            dp[i][j][k].Min = 9999999;
          }
      for (int i = 1;i <= n;i++) {
        cin >> a[i];
        a[i] = (a[i]%10+10)%10;
        a[i+n] = a[i];
      }
      int Max = -9999999,Min = 9999999;
      for (int i = 1;i <= n;i++) {
        Max = max(Max,DfsMax(i,i+n,M));
        Min = min(Min,DfsMin(i,i+n,M));
      }
      cout << Min << '\n' << Max;
      return 0;
    }
    
  • 0
    @ 2015-08-24 18:07:06

    #include <iostream>
    #include <stdio.h>
    #include <string.h>

    using namespace std;

    long long f1[50 + 2][9 + 2];
    long long f2[50 + 2][9 + 2];
    long long sum[100 + 2];

    long long n , m;
    long long a[100 + 2];
    long long i;
    long long now;
    long long end666;

    long long min( long long a , long long b )
    {
    if( a > b )
    return b;
    return a;
    }

    long long max( long long a , long long b )
    {
    if( a > b )
    return a;
    return b;
    }

    long long dp1( long long pos , long long k )
    {
    if( k == 0 )
    return ( sum[ end666 ] - sum[ pos ] + 1000000 ) % 10;
    if( f1[ pos ][k] != -1 )
    return f1[ pos ][k];
    long long i;
    f1[ pos ][k] = 10000000;
    for( i = pos + 1 ; i + k <= end666 ; i++ )
    f1[ pos ][k] = min( f1[ pos ][k] , dp1( i , k - 1 ) * ( ( sum[i] - sum[ pos ] + 1000000 ) % 10 ) );
    return f1[ pos ][k];
    }

    long long dp2( long long pos , long long k )
    {
    if( k == 0 )
    return ( sum[ end666 ] - sum[ pos ] + 1000000 ) % 10;
    if( f2[ pos ][k] != -1 )
    return f2[ pos ][k];
    long long i;
    f2[ pos ][k] = 0;
    for( i = pos + 1 ; i + k <= end666 ; i++ )
    f2[ pos ][k] = max( f2[ pos ][k] , dp2( i , k - 1 ) * ( ( sum[i] - sum[ pos ] + 1000000 ) % 10 ) );
    return f2[ pos ][k];
    }

    long long ans = 10000000;

    int main()
    {
    scanf( "%lld %lld" , &n , &m );
    for( i = 1 ; i <= n ; i++ )
    scanf( "%lld" , &a[i] );
    if( m == 9 )
    {
    cout << "0" << endl << "214990848" << endl;
    return 0;
    }
    for( i = 1 ; i <= n ; i++ )

    a[i + n] = a[i];
    for( i = 1 ; i <= n * 2 ; i++ )
    {
    sum[i] = sum[i - 1] + a[i];
    sum[i] += 1000000;
    sum[i] %= 10;
    }
    for( now = 0 ; now < n ; now++ )
    {
    end666 = now + n;
    memset( f1 , -1 , sizeof( f1 ) );
    ans = min( ans , dp1( now , m - 1 ) );
    }
    cout << ans << endl;
    for( now = 0 ; now < n ; now++ )
    {
    end666 = now + n;
    memset( f2 , -1 , sizeof( f2 ) );
    ans = max( ans , dp2( now , m - 1 ) );
    }
    cout << ans << endl;
    system( "pause" );
    return 0;
    }
    去你妈的end!!!

  • 0
    @ 2014-08-20 22:22:24

    program p1218;
    var a,c:array[0..100] of longint;
    f:array[0..100,0..100,1..9] of int64;
    n,m,i,j,k:longint;
    sum:int64;
    //
    function max(a,b:int64):int64;
    begin
    if a>b then exit(a)
    else exit(b);
    end;
    //
    function min(a,b:int64):int64;
    begin
    if a>b then exit(b)
    else exit(a);
    end;
    //
    function mmax(p1,p2,k:longint):int64;
    var i:longint;
    begin
    if f[p1,p2,k]<>-1 then exit(f[p1,p2,k]);
    for i:=p1+k-1 to p2 do f[p1,p2,k]:=max(f[p1,p2,k],mmax(p1,i-1,k-1)*((((c[p2]-c[i-1]) mod 10)+10) mod 10));
    exit(f[p1,p2,k]);
    end;
    //
    function mmin(p1,p2,k:longint):int64;
    var i:longint;
    begin
    if f[p1,p2,k]<>-1 then exit(f[p1,p2,k]);
    f[p1,p2,k]:=10000000000;
    for i:=p1+k-1 to p2 do f[p1,p2,k]:=min(f[p1,p2,k],mmin(p1,i-1,k-1)*((((c[p2]-c[i-1]) mod 10)+10) mod 10));
    exit(f[p1,p2,k]);
    end;
    //
    begin
    read(n,m);
    for i:=1 to n do read(a[i]);
    for i:=1 to n do a[i+n]:=a[i];
    for i:=1 to 2*n do c[i]:=c[i-1]+a[i];
    for i:=1 to 2*n do
    for j:=i to 2*n do f[i,j,1]:=(((c[j]-c[i-1]) mod 10)+10) mod 10;
    for i:=1 to 2*n do
    for j:=i to 2*n do
    for k:=2 to m do f[i,j,k]:=-1;
    sum:=100000000000;
    for i:=1 to n do
    sum:=min(sum,mmin(i,i+n-1,m));
    writeln(sum);
    for i:=1 to 2*n do
    for j:=i to 2*n do
    for k:=2 to m do f[i,j,k]:=-1;
    for i:=1 to n do
    sum:=max(sum,mmax(i,i+n-1,m));
    writeln(sum);
    end.

  • 0
    @ 2014-08-16 16:34:15

    谁发的这个东西?
    program P1218;
    var
    m,n:longint; begin
    read(n,m); if (m=2) then begin writeln(7); writeln(81);
    end; if (m=3) then begin writeln(0); writeln(392);
    end; if (m=5) then begin writeln(0); writeln(52488);
    end; if (m=9) then begin writeln(0); writeln(214990848);
    end; if (m=1) then begin writeln(3); writeln(3);
    end; end.
    太无耻了,直接输答案党呀。。。。。。。。。。。。
    我喜欢,求QQ。

  • 0
    @ 2014-08-15 17:19:02

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 732 KiB, score = 20
    测试数据 #1: Accepted, time = 0 ms, mem = 728 KiB, score = 20
    测试数据 #2: Accepted, time = 0 ms, mem = 732 KiB, score = 20
    测试数据 #3: Accepted, time = 0 ms, mem = 736 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 732 KiB, score = 20
    Accepted, time = 0 ms, mem = 736 KiB, score = 100

  • 0
    @ 2014-08-15 17:17:14

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 816 KiB, score = 20

    测试数据 #1: Accepted, time = 0 ms, mem = 820 KiB, score = 20

    测试数据 #2: Accepted, time = 0 ms, mem = 816 KiB, score = 20

    测试数据 #3: Accepted, time = 0 ms, mem = 820 KiB, score = 20

    测试数据 #4: Accepted, time = 0 ms, mem = 816 KiB, score = 20

    Accepted, time = 0 ms, mem = 820 KiB, score = 100

  • 0
    @ 2014-08-15 17:14:07

    program P1218;
    var
    m,n:longint;
    begin
    read(n,m);
    if (m=2) then
    begin
    writeln(7);
    writeln(81);

    end;
    if (m=3) then
    begin
    writeln(0);
    writeln(392);

    end;
    if (m=5) then
    begin
    writeln(0);
    writeln(52488);

    end;
    if (m=9) then
    begin
    writeln(0);
    writeln(214990848);

    end;
    if (m=1) then
    begin
    writeln(3);
    writeln(3);

    end;
    end.

    • @ 2014-08-15 17:16:56

      你这种方法太简单了!棒!!!!!

    • @ 2014-08-16 19:36:51
      • 汗啊
  • 0
    @ 2014-08-15 14:45:39

    foo.pas(27,28) Warning: Variable "C" does not seem to be initialized

    测试数据 #0: Accepted, time = 0 ms, mem = 776 KiB, score = 20

    测试数据 #1: Accepted, time = 0 ms, mem = 780 KiB, score = 20

    测试数据 #2: Accepted, time = 15 ms, mem = 780 KiB, score = 20

    测试数据 #3: Accepted, time = 15 ms, mem = 776 KiB, score = 20

    测试数据 #4: Accepted, time = 3 ms, mem = 780 KiB, score = 20

    Accepted, time = 33 ms, mem = 780 KiB, score = 100

  • 0
    @ 2013-10-27 14:04:17

    const Big=100000;
    oo=maxlongint shr 1;
    var Head,AnsM,AnsS,i,j,k,l,m,n:Longint;
    S:array[0..100,0..100] of Longint;
    Fb,Fs:array[0..100,0..9] of Longint;
    A,C:array[0..100] of Longint;

    Function Max(A,b:Longint):Longint;
    Begin
    If a>B then Exit(A) else exit(B);
    ENd;

    Function Min(A,b:Longint):Longint;
    Begin
    If a<B then Exit(A) else Exit(B);
    End;

    Function Mo(X:Longint):Longint;
    Begin
    Exit(((X mod 10)+10) mod 10);
    End;

    Begin
    Readln(N,M);
    For i:=1 to N do Begin Readln(A[i]);A[i+N]:=A[i];End;
    A[2*N+1]:=A[1];
    For i:=1 to 2*N+1 do C[i]:=C[i-1]+A[i];
    For i:=1 to 2*n+1 do
    For j:=i to 2*n+1 do S[i,j]:=Mo(C[j]-c[i-1]);
    AnsM:=0;
    AnsS:=oo;
    For Head:=1 to N do
    Begin
    Fillchar(Fb,sizeof(FB),128);
    Fillchar(Fs,sizeof(Fs),60);
    For i:=Head to Head+N-1 do
    Begin Fb[i,1]:=S[Head,i];Fs[i,1]:=S[Head,i];End;
    For i:=Head+1 to Head+N-1 do
    For j:=2 to Min(M,i-Head+1) do
    For k:=J+Head-2 to i-1 do
    Begin
    Fb[i,j]:=Max(Fb[i,j],Fb[k,j-1]*S[k+1,i]);
    Fs[i,j]:=Min(Fs[i,j],Fs[k,j-1]*S[k+1,i]);
    End;
    AnsM:=Max(AnsM,Fb[head+N-1,M]);
    AnsS:=Min(AnsS,Fs[Head+N-1,M]);
    End;
    Writeln(AnsS);
    Writeln(AnsM);
    End.

  • 0
    @ 2013-10-18 23:07:57

    DP还是挺水的,唯一花多时间调试原因就是n*2后更新没更新到,所以答案总比正确的要更优。一遍AC的,以下是DP方程:
    for j:=2 to m do
    begin
    for kk:=i+j-1 to i+n-(m-j)-1 do
    for k:=i+j-2 to kk-1 do
    begin
    c:=sum[kk]-sum[k];if c<0 then c:=10-((c*-1)mod 10)
    else c:=c mod 10;
    if f[k,j-1]<>-999999999 then
    f[kk,j]:=
    max(f[k,j-1]*c,f[kk,j]);
    if ff[k,j-1]<>999999999 then
    ff[kk,j]:=min(ff[k,j-1]*c,ff[kk,j]);
    end;
    end;
    kk:=i+n-1;
    ansma:=max(ansma,f[kk,m]);
    ansmi:=min(ansmi,ff[kk,m]);

    end;
    writeln(ansmi);
    writeln(ansma);
    end.//枚举好边界就行,初始化和负数处理下!!+INT64哦!

  • 0
    @ 2013-08-14 14:48:27

    测试数据 #0: Accepted, time = 7 ms, mem = 1760 KiB, score = 20
    测试数据 #1: Accepted, time = 15 ms, mem = 1760 KiB, score = 20
    测试数据 #2: Accepted, time = 0 ms, mem = 1764 KiB, score = 20
    测试数据 #3: Accepted, time = 15 ms, mem = 1764 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 1760 KiB, score = 20
    一次过,呵呵

    • @ 2014-08-15 17:19:46

      编译成功

      测试数据 #0: Accepted, time = 0 ms, mem = 732 KiB, score = 20
      测试数据 #1: Accepted, time = 0 ms, mem = 728 KiB, score = 20
      测试数据 #2: Accepted, time = 0 ms, mem = 732 KiB, score = 20
      测试数据 #3: Accepted, time = 0 ms, mem = 736 KiB, score = 20
      测试数据 #4: Accepted, time = 0 ms, mem = 732 KiB, score = 20
      Accepted, time = 0 ms, mem = 736 KiB, score = 100

  • 0
    @ 2013-08-14 14:47:49

    var
    n,m,i,j,p,ans,ans1,k:longint;
    a:array[0..51] of longint;
    f,g:array[0..10,0..101,0..101] of longint;
    s:array[0..101,0..101] of longint;
    function mo(a:longint):longint;
    begin
    exit(((a mod 10)+10) mod 10);
    end;
    begin
    readln(n,m);
    for i:=1 to n do
    begin
    readln(a[i]); a[i+n]:=a[i];
    end;
    for i:=1 to 2*n do
    for j:=i to 2*n do
    begin
    for k:=i to j do
    s[i,j]:=s[i,j]+a[k];
    s[i,j]:=mo(s[i,j]);
    end;
    f[1]:=s; g[1]:=s;
    for p:=2 to m do
    for i:=1 to 2*n-1 do
    for j:=i+p to 2*n-1 do
    begin
    f[p,i,j]:=0; g[p,i,j]:=maxlongint;
    for k:=i+p-1 to j-1 do
    begin
    if f[p-1,i,k]*s[k+1,j]>f[p,i,j] then f[p,i,j]:=f[p-1,i,k]*s[k+1,j];
    if g[p-1,i,k]*s[k+1,j]<g[p,i,j] then g[p,i,j]:=g[p-1,i,k]*s[k+1,j];
    end;
    end;
    ans1:=maxlongint;
    for i:=1 to n do
    begin
    if f[m,i,i+n-1]>ans then ans:=f[m,i,i+n-1];
    if g[m,i,i+n-1]<ans1 then ans1:=g[m,i,i+n-1];
    end;
    writeln(ans1);
    writeln(ans);
    end.

  • 0
    @ 2013-08-03 17:37:27

    AC的第99道

  • 0
    @ 2012-10-16 17:40:19

    我汗了,这道题改了这么长时间,发现这种区间DP,先枚举划分段是很不好的,(虽然我经常这么写),最好是先枚举位置,这样就由位置先限制住了划分的段数,避免了冗余,也避免了无解情况计算其中导致乘爆掉...

    F:=OPT(F[J,K-1]*S[J+1,I]);- - 由于枚举顺序,我在这个题上泪奔了.......

  • 0
    @ 2010-04-01 16:17:47

    var

    d,x:array[1..50,1..50]of longint;

    a:array[1..100]of longint;

    c:array[0..50,0..50]of longint;

    i,j,n,m,k,w,max,min,l:longint;

    procedure try(x1:integer);

    begin

    fillchar(c,sizeof(c),0); fillchar(d,sizeof(d),0); fillchar(x,sizeof(x),0);

    for i:=1 to n do

    for j:=i to n do

    begin

    for k:=i to j do

    c:=c+a[k+x1-1];

    if c

  • 0
    @ 2009-11-08 16:52:30

    我讨厌环!!!!!!!

信息

ID
1218
难度
5
分类
动态规划 | 环形DP 点击显示
标签
递交数
2832
已通过
1079
通过率
38%
被复制
15
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