题解

47 条题解

  • 1
    @ 2019-03-13 20:29:33

    被两点坑了
    1.最大=最大*最大or最小*最小;最小=最大*最大or最大*最小or最小*最大or最大*最小
    2.初始化不能是0。

    #include<cstdio>
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    using namespace std;
    const long long inf = 0x7fffffffffffffff;
    char c[110];
    int vis[110][110][2],n;  //3-d 0小 1大
    long long d[110][110][2],smax,smin=inf;
    long long dp(int i,int j,int f){ //左闭右开
        long long &ans = d[i][j][f];
        if(vis[i][j][f]) return ans;
        vis[i][j][f] = 1;
        if(i+1==j) return ans;
        for(int k=i+1;k<j;k++){
            if(f) ans = max(ans,c[k-1]=='*'?max(dp(i,k,1)*dp(k,j,1),dp(i,k,0)*dp(k,j,0)):dp(i,k,1)+dp(k,j,1));
            else ans = min(ans,c[k-1]=='*'?min(min(dp(i,k,0)*dp(k,j,0),dp(i,k,1)*dp(k,j,1)),min(dp(i,k,1)*dp(k,j,0),dp(i,k,0)*dp(k,j,1))):dp(i,k,f)+dp(k,j,f));
        }
        return ans;
    }
    int main(){
        ios::sync_with_stdio(false);
        cin>>n;
        for(int i=0;i<n;i++){
            cin>>c[i];
            c[n+i] = c[i];
        }
        for(int i=0;i<2*n;i++) for(int j=0;j<2*n;j++) d[i][j][0] = inf,d[i][j][1]=-inf;
        for(int i=0;i<n;i++){
            cin>>d[i][i+1][0];
            d[i+n][i+n+1][1] = d[i+n][i+n+1][0] = d[i][i+1][1] = d[i][i+1][0];
        }
        for(int i=0;i<n;i++){
            smax = max(smax,dp(i,i+n,1));
            smin = min(smin,dp(i,i+n,0));
        }
        cout<<smax<<'\n'<<smin;
        return 0;
    }
    
    
  • 0
    @ 2017-08-22 00:13:29

    由于乘法的特殊性,状态转移需分5种情况,然后是简单的区间dp了

    代码如下

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    using namespace std;
    int n,dp[104][104][2],maxx=-0x3f3f3f3f,minn=0x3f3f3f3f;//0---max,1---min
    string s;
    int work(int a,int b,char ch){
        if(ch=='+') return a+b;
        else return a*b;
    }
    int main(){
        scanf("%d",&n);
        string s;
        cin>>s;
        s=s+s;
        for(int i=1;i<2*n;i++)
            for(int j=1;j<2*n;j++){
                dp[i][j][1]=0x3f3f3f3f;
                dp[i][j][0]=-0x3f3f3f3f;
            }
        for(int i=1;i<=n;i++){
            scanf("%d",&dp[i][i][0]);
            dp[n+i][n+i][1]=dp[n+i][n+i][0]=dp[i][i][1]=dp[i][i][0];
        }
        for(int l=1;l<=n;l++){
            for(int i=1;i<=2*n-1;i++){
                int j=i+l-1;
                if(j>=2*n) break;
                for(int k=i;k<j;k++){
                    dp[i][j][0]=max(dp[i][j][0],work(dp[i][k][0],dp[k+1][j][0],s[k-1]));
                    dp[i][j][0]=max(dp[i][j][0],work(dp[i][k][1],dp[k+1][j][1],s[k-1]));
                    dp[i][j][1]=min(dp[i][j][1],work(dp[i][k][1],dp[k+1][j][1],s[k-1]));
                    dp[i][j][1]=min(dp[i][j][1],work(dp[i][k][0],dp[k+1][j][1],s[k-1]));
                    dp[i][j][1]=min(dp[i][j][1],work(dp[i][k][1],dp[k+1][j][0],s[k-1]));
                }
            }
        }
        for(int i=1;i<=n;i++){
            maxx=max(maxx,dp[i][i+n-1][0]);
            minn=min(minn,dp[i][i+n-1][1]);
        }
        printf("%d\n%d",maxx,minn);
    }
    
  • 0
    @ 2017-02-26 19:58:10

    总算AC了其实并不难= =

  • 0
    @ 2016-08-02 21:48:37

    强烈BS此题,有负数的情况!

  • 0
    @ 2016-05-22 11:04:21
    1. 破环成链
    2. 注意初始化:不能赋0
    3. 注意数据范围:longint,保险开int64
    4. 注意**负数**
      1. 最大值可以由*最大x最大*和*最小x最小*(负负得正)两种情况得到
      2. 最小值可以由*最小x最小*和*最大x最小*(正负得负)得到
    uses math;
    const inf=99999999999;
    var
      n,
      amax,amin,
      i,j,k,l:longint; //[i,j],j-i+1=l,k in [i,j]
      e:array[0..100] of char;
      v:array[0..100] of int64;
      dmax,dmin:array[0..100,0..100] of int64;
    
    function union(a,b:longint;c:char):longint;
    begin
      case c of
        '*':union:=a*b;
        '+':union:=a+b;
      end;
    end;
    
    begin
      readln(n);
      for i:=1 to n do begin
        read(e[i]);
        e[i+n]:=e[i];
      end;
      for i:=1 to n do begin
        read(v[i]);
        v[i+n]:=v[i];
      end;
      fillqword(dmax,sizeof(dmax) div 8,-inf);
      fillqword(dmin,sizeof(dmin) div 8,inf);
      for i:=1 to 2*n do begin
        dmax[i,i]:=v[i];
        dmin[i,i]:=v[i];
      end;
    
      for l:=2 to 2*n do begin
        for i:=1 to 2*n+1-l do begin
          j:=i+l-1;
          for k:=i to j-1 do begin
            dmin[i,j]:=min(min(dmin[i,j],
                               union(dmax[i,k],dmin[k+1,j],e[k])),
                           min(union(dmin[i,k],dmax[k+1,j],e[k]),
                               union(dmin[i,k],dmin[k+1,j],e[k]))
                           );
            dmax[i,j]:=max(dmax[i,j],
                           max(union(dmax[i,k],dmax[k+1,j],e[k]),
                               union(dmin[i,k],dmin[k+1,j],e[k]))
                           );
          end;
        end;
      end;
    
      amax:=-inf;amin:=inf;
      for i:=1 to n do begin
        amax:=max(amax,dmax[i,i+n-1]);
        amin:=min(amin,dmin[i,i+n-1]);
      end;
      writeln(amax);writeln(amin);
    end.
    
  • 0
    @ 2015-09-28 20:28:35

    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>

    using namespace std;

    int n;
    int fmin0[105][105], fmax0[105][105];
    char prtr[105];
    bool vis[105][105], vis1[105][105];

    int calcu(int x0, int x1, char c)
    {
    if (c == '*')
    return x0 * x1;
    else
    return x0 + x1;
    }
    int max0(int x0, int x1, int x2, int x3, int x4)
    {
    return max(x4, max(x0, max(x1, max(x2, x3))));
    }
    int min0(int x0, int x1, int x2, int x3, int x4)
    {
    return min(x4, min(x0, min(x1, min(x2, x3))));
    }
    int main(int argc, const char *argv[])
    {
    scanf("%d", &n);
    scanf("%s", prtr);
    for (int i = 1; i <= n * 2; ++i)
    for (int j = 1; j <= n * 2; ++j)
    {
    fmin0[i][j] = 2147483647;
    fmax0[i][j] = -2147483647;
    }
    for (int i = 1; i <= n; ++i)
    {
    scanf("%d", &fmin0[i][i]);
    fmin0[i + n][i + n] = fmax0[i + n][i + n] = fmax0[i][i] = fmin0[i][i];
    }
    for (int i = n; i >= 1; --i)
    prtr[i] = prtr[i - 1];
    for (int i = n + 1; i <= n * 2; ++i)
    prtr[i] = prtr[i - n];
    for (int p = 1; p <= n - 1; ++p)
    for (int i = 1; i <= n * 2 - p; ++i)
    for (int j = i + 1; j <= i + p; ++j)
    {
    if (vis[i][j])
    continue;
    vis[i][j] = true;
    for (int k = i; k < j; ++k)
    {
    fmax0[i][j] = max0(
    fmax0[i][j],
    calcu(fmax0[i][k], fmax0[k + 1][j], prtr[k]),
    calcu(fmax0[i][k], fmin0[k + 1][j], prtr[k]),
    calcu(fmin0[i][k], fmin0[k + 1][j], prtr[k]),
    calcu(fmin0[i][k], fmax0[k + 1][j], prtr[k])
    );
    fmin0[i][j] = min0(
    fmin0[i][j],
    calcu(fmax0[i][k], fmax0[k + 1][j], prtr[k]),
    calcu(fmax0[i][k], fmin0[k + 1][j], prtr[k]),
    calcu(fmin0[i][k], fmin0[k + 1][j], prtr[k]),
    calcu(fmin0[i][k], fmax0[k + 1][j], prtr[k])
    );
    }
    }
    int ans1, ans2;
    ans1 = -2147483647;
    ans2 = 2147483647;
    for (int i = 1; i <= n; ++i)
    {
    ans1 = max(ans1, fmax0[i][i + n - 1]);
    ans2 = min(ans2, fmin0[i][i + n - 1]);
    }
    printf("%d\n%d\n", ans1, ans2);
    return 0;
    }

  • 0
    @ 2015-01-11 22:34:43

    发个清晰一点的代码。虽然比较长,但应该还好理解。说明几点:
    1. dbNum=2*num-1,用于把数据展开(如1 1 3 4展开为1 1 3 4 1 1 3)从而规避复杂的取余处理
    2. 状态转移时,要区分加和乘。乘法会有5种情况
    3. 函数MAX和MIN有5个参数,当比较的变量数小于5时,使用极值填充来避免干扰

    #include <stdio.h>
    #include <limits.h>
    char operators[105];
    int operands[105];
    int maxArr[105][105],minArr[105][105];
    int MAX(int a,int b,int c,int d,int e){
    a=a>b?a:b;
    a=a>c?a:c;
    a=a>d?a:d;
    return a>e?a:e;
    }
    int MIN(int a,int b,int c,int d,int e){
    a=a<b?a:b;
    a=a<c?a:c;
    a=a<d?a:d;
    return a<e?a:e;
    }
    int main(){
    int num,dbNum;
    int i,begin,end,mid;
    scanf("%d",&num);
    scanf("%s",operators);
    for(i=0;i<num;i++)
    scanf("%d",&operands[i]);
    dbNum=num*2-1;
    for(i=num;i<dbNum;i++){
    operators[i]=operators[i-num];
    operands[i]=operands[i-num];
    }
    for(begin=0;begin<dbNum;begin++){
    for(end=0;end<dbNum;end++){
    maxArr[begin][end]=LONG_MIN;
    minArr[begin][end]=LONG_MAX;
    }
    }
    for(i=0;i<dbNum;i++)
    maxArr[i][i]=minArr[i][i]=operands[i];
    for(i=1;i<num;i++){
    for(begin=0;begin<dbNum;begin++){
    end=begin+i;
    if(end>dbNum)
    break;
    for(mid=begin;mid<end;mid++){
    if(operators[mid]=='+'){
    maxArr[begin][end]=MAX(
    maxArr[begin][end],
    maxArr[begin][mid]+maxArr[mid+1][end],
    LONG_MIN,LONG_MIN,LONG_MIN
    );
    minArr[begin][end]=MIN(
    minArr[begin][end],
    minArr[begin][mid]+minArr[mid+1][end],
    LONG_MAX,LONG_MAX,LONG_MAX
    );
    }else if(operators[mid]=='*'){
    maxArr[begin][end]=MAX(
    maxArr[begin][end],
    maxArr[begin][mid]*maxArr[mid+1][end],
    maxArr[begin][mid]*minArr[mid+1][end],
    minArr[begin][mid]*maxArr[mid+1][end],
    minArr[begin][mid]*minArr[mid+1][end]
    );
    minArr[begin][end]=MIN(
    minArr[begin][end],
    maxArr[begin][mid]*maxArr[mid+1][end],
    maxArr[begin][mid]*minArr[mid+1][end],
    minArr[begin][mid]*maxArr[mid+1][end],
    minArr[begin][mid]*minArr[mid+1][end]
    );
    }
    }
    }
    }
    int n=LONG_MIN,m=LONG_MAX;
    for(begin=0;begin<num;begin++){
    if(n<maxArr[begin][begin+num-1])
    n=maxArr[begin][begin+num-1];
    if(m>minArr[begin][begin+num-1])
    m=minArr[begin][begin+num-1];
    }
    printf("%d\n%d\n",n,m);
    return 0;
    }

  • 0
    @ 2014-08-15 08:40:07

    *program P1565;
    *var
    * minopt,maxopt:array[0..200,0..200] of longint;
    * ch:array[1..200] of char;
    * point:array[1..200] of longint;
    * n,i,k,l,j,mi,ma:longint;
    *function min(x,y:longint):longint;
    * begin
    * if x>y then exit(y);
    * exit(x);
    * end;
    *function max(x,y:longint):longint;
    * begin
    * max:=x;
    * if x<y then max:=y;
    * end;
    *begin
    * readln(n);
    * for i:=1 to n do
    * begin
    * read(ch[i]);
    * ch[i+n]:=ch[i];
    * end;
    * for i:=1 to n do
    * begin
    * read(point[i]);
    * point[i+n]:=point[i];
    * minopt[i,i]:=point[i];
    *minopt[i+n,i+n]:=point[i];
    * maxopt[i,i]:=point[i];
    * maxopt[i+n,i+n]:=point[i];
    *end;
    * for i:=1 to 2*n do
    * for j:=1 to 2*n do
    * if i<>j then
    * begin
    * maxopt[i,j]:=-maxlongint;
    * minopt[i,j]:=maxlongint;
    * end;
    * for l:=1 to n-1 do
    * for i:=1 to 2*n-1-l do
    * begin
    * for k:=i to i+l-1 do
    * begin
    * if ch[k]='+' then
    * begin
    * maxopt[i,i+l]:=max(maxopt[i,i+l],maxopt[i,k]+maxopt[k+1,i+l]);
    * minopt[i,i+l]:=min(minopt[i,i+l],minopt[i,k]+minopt[k+1,i+l]);
    * end;
    * if ch[k]='*' then
    * begin
    * maxopt[i,i+l]:=max(maxopt[i,i+l],maxopt[i,k]*maxopt[k+1,i+l]);
    * minopt[i,i+l]:=min(minopt[i,i+l],minopt[i,k]*minopt[k+1,i+l]);
    * maxopt[i,i+l]:=max(maxopt[i,i+l],minopt[i,k]*minopt[k+1,i+l]);
    * minopt[i,i+l]:=min(minopt[i,i+l],maxopt[i,k]*maxopt[k+1,i+l]);
    * maxopt[i,i+l]:=max(maxopt[i,i+l],minopt[i,k]*maxopt[k+1,i+l]);
    * minopt[i,i+l]:=min(minopt[i,i+l],maxopt[i,k]*minopt[k+1,i+l]);
    * maxopt[i,i+l]:=max(maxopt[i,i+l],maxopt[i,k]*minopt[k+1,i+l]);
    * minopt[i,i+l]:=min(minopt[i,i+l],minopt[i,k]*maxopt[k+1,i+l]);
    * end;
    * end;
    * end;
    *mi:=maxlongint;
    * ma:=-maxlongint;
    * for i:=1 to n do
    * begin
    * if mi>minopt[i,i+n-1] then mi:=minopt[i,i+n-1];
    * if ma<maxopt[i,i+n-1] then ma:=maxopt[i,i+n-1];
    * end;
    * writeln(ma);
    * writeln(mi);
    *end.

  • 0
    @ 2014-08-07 09:01:30

    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <cmath>
    using namespace std;
    const long long maxe=1000000000;
    #define ll long long
    ll minn,maxn,save[1000],f[200][200][2];
    ll Min(ll a,ll b,ll c,ll d)
    {
    ll tmp=maxe;
    tmp=min(tmp,a);
    tmp=min(tmp,b);
    tmp=min(tmp,c);
    tmp=min(tmp,d);
    return tmp;
    }
    ll Max(ll a,ll b,ll c,ll d)
    {
    ll tmp=-maxe;
    tmp=max(tmp,a);
    tmp=max(tmp,b);
    tmp=max(tmp,c);
    tmp=max(tmp,d);
    return tmp;
    }
    char st[200];
    int n,op[200];
    void dfs(int l,int r)
    {
    ll num=maxe;
    ll ans=-maxe;
    if(f[l][r][0]==-maxe)
    {
    if(l==r)
    {
    f[l][r][0]=save[l];
    }
    else
    {
    for(int i=l;i<r;i++)
    {
    dfs(l,i);
    dfs(i+1,r);
    if(op[i]==1)
    {
    num=f[l][i][0]+f[i+1][r][0];
    }
    if(op[i]==2)
    {
    num=Max(f[l][i][0]*f[i+1][r][0],f[l][i][1]*f[i+1][r][0],f[l][i][0]*f[i+1][r][1],f[l][i][1]*f[i+1][r][1]);
    }
    if (num>ans)
    ans=num;
    }
    f[l][r][0]=ans;
    }
    }
    ans=maxe;
    if(f[l][r][1]==maxe)
    {
    if(l==r)
    {
    f[l][r][1]=save[l];
    }
    else
    {
    for(int i=l;i<r;i++)
    {
    dfs(l,i);
    dfs(i+1,r);
    if(op[i]==1)
    {
    num=f[l][i][1]+f[i+1][r][1];
    }
    if(op[i]==2)
    {
    num=Min(f[l][i][0]*f[i+1][r][0],f[l][i][1]*f[i+1][r][0],f[l][i][0]*f[i+1][r][1],f[l][i][1]*f[i+1][r][1]);
    }
    if (num<ans)
    ans=num;
    }
    f[l][r][1]=ans;
    }
    }
    }
    int main()
    {
    minn=maxe;
    maxn=-maxe;
    scanf("%d",&n);
    scanf("%s",st+1);
    for(int i=1;i<=2*n;i++)
    {
    for(int j=1;j<=2*n;j++)
    {
    f[i][j][1]=maxe;
    f[i][j][0]=-maxe;
    }
    }
    for(int i=1;i<=n;i++)
    {
    if(st[i]=='+')
    op[i]=1;
    if(st[i]=='*')
    op[i]=2;
    scanf("%lld",&save[i]);
    }
    for(int i=n+1;i<=2*n;i++)
    {
    op[i]=op[i-n];
    save[i]=save[i-n];
    }
    ll ans1=minn;
    ll ans2=maxn;
    for(int i=1;i<=n;i++)
    {
    dfs(i,i+n-1);
    maxn=max(maxn,f[i][i+n-1][0]);
    minn=min(minn,f[i][i+n-1][1]);
    }
    printf("%lld\n%lld",maxn,minn);
    return 0;
    }

  • 0
    @ 2014-08-07 09:01:23

    测试数据 #0: Accepted, time = 0 ms, mem = 876 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 880 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 876 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 880 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 876 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 876 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 876 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 876 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 880 KiB, score = 10
    测试数据 #9: Accepted, time = 15 ms, mem = 880 KiB, score = 10
    Accepted, time = 30 ms, mem = 880 KiB, score = 100

  • 0
    @ 2009-10-06 20:40:34

    可恶!!!!辛辛苦苦写的高精居然不用。。。。。。而且题目没说明有负数!也没说数的大小范围!!!!

    简直就是个三无题目!!!!BS!!!!!!!!坑人啊!!!

  • 0
    @ 2009-10-06 07:33:18

    数组初值不要设成0,要设成max或min

  • 0
    @ 2009-09-05 22:28:04

    有点像石子合并的DP

  • 0
    @ 2009-08-26 12:46:33

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:0ms

    真不容易,提交了这么多次,首先要注意负数乘法的问题,如果只有第九个点不过,就象我很多次没过,就是max[]数组初值的问题

    memset(max,-1,sizeof(max));

    memset(min,1,sizeof(min));

    就可以了,感谢wjh……

  • 0
    @ 2009-08-06 09:48:04

    奇怪!怎么没说顶点值的范围,longint够了吗?要开long long吗?要上高精度吗?出题真不严谨!!

  • 0
    @ 2009-08-03 14:17:45

    const Max_size=10000000;

    var work:array[0..100] of boolean;

    fmax,fmin:array[0..100,0..100] of longint;

    a:array[0..100] of longint;

    n,m,i,k,j,l,p,maxans,minans:longint;

    ch:char;

    function max(a,b,c,d,e:longint):longint;

    begin

    max:=a;

    if maxe then min:=e;

    end;

    begin

    readln(n);

    for i:=1 to n do

    begin

    read(ch); work[i]:=ch='+';

    end;

    readln;

    for i:=1 to n do read(a[i]);

    m:=n*2-1;

    for i:=n+1 to m do

    begin

    work[i]:=work; a[i]:=a;

    end;

    fillchar(fmax,sizeof(fmax),200);

    fillchar(fmin,sizeof(fmin),100);

    for i:=1 to m do

    begin

    fmax:=a[i]; fmin:=a[i];

    end;

    for i:=1 to n-1 do

    for k:=1 to m do

    begin

    j:=i+k; if j>m then break;

    for p:=k to j-1 do

    if work[p] then

    begin

    fmax[k,j]:=max(fmax[k,j],fmax[k,p]+fmax[p+1,j],-Max_size,-Max_size,-Max_size);

    fmin[k,j]:=min(fmin[k,j],fmin[k,p]+fmin[p+1,j],Max_size,Max_size,Max_size);

    end else

    begin

    fmax[k,j]:=max(fmax[k,j],fmin[k,p]*fmin[p+1,j],fmax[k,p]*fmax[p+1,j],fmax[k,p]*fmin[p+1,j],fmin[k,p]*fmax[p+1,j]);

    fmin[k,j]:=min(fmin[k,j],fmin[k,p]*fmin[p+1,j],fmax[k,p]*fmax[p+1,j],fmax[k,p]*fmin[p+1,j],fmin[k,p]*fmax[p+1,j]);

    end;

    end;

    maxans:=-Max_size; minans:=Max_size;

    for i:=1 to n do

    begin

    if fmax>maxans then maxans:=fmax;

    if fmin

  • 0
    @ 2009-08-01 19:30:22

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 25ms

    ├ 测试数据 09:答案正确... 9ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:34ms

    终于过了,好BT啊

  • 0
    @ 2009-07-25 11:12:49

    搞了N久,比大小比错了,晕死,AC率又降了,哎,细节啊!!!!我是第111个AC的!

    庆祝第200题!!!

    Flag   Accepted

    题号   P1565

    类型(?)   动态规划

    通过   111人

    提交   553次

    通过率   20%

    难度   1

  • 0
    @ 2009-07-25 11:09:55

    .............

  • 0
    @ 2009-07-16 19:13:20

    zgx我恨你。。。。

信息

ID
1565
难度
7
分类
动态规划 | 环形DP 点击显示
标签
递交数
1706
已通过
319
通过率
19%
被复制
3
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