1.方程

给出非负整数n，统计不定方程x+y^2+z^3=n,的非负整数解（x，y，z）的数量。

令k=n-z^3;

则k最多有10^6个值

x+y^2=k

y的取值个数有sqrt(k)+1个

累加y

 

#include<fstream>
#include<cmath>
using namespace std;
ifstream cin("equation.in");
ofstream cout("equation.out");
long long n,k,ans;
long long x,y,z;
int main()
{
    cin>>n;
    ans=0;
    z=0;
    while ((z*z*z)<=n)
    {
          k=n-z*z*z;
          y=(int)sqrt((double)k)+1;
          ans+=y;
          z++;
    }
    cout<<ans;
    return 0;
}

2.方程式

 

模拟，字符串处理

注意结果是0的时候

 

代码 
var s,left,right,temp:string;
    tot,num,t,tmp,i:longint;
    flag:boolean;
    ans:double;
begin
  assign(input,'equationagain.in');reset(input);
  assign(output,'equationagain.out');rewrite(output);
  readln(s);
  t:=pos('=',s);
  left:=copy(s,1,t-1);
  right:=copy(s,t+1,length(s)-t);
  t:=length(left);
  for i:=1 to t do
    begin
      if left[i] in ['0'..'9'] then temp:=temp+left[i]
        else
          begin
            if left[i]='x' then
              begin
                if temp='' then tmp:=1 else val(temp,tmp);
                if flag then tmp:=-tmp;
                flag:=false;
                inc(tot,tmp);
                temp:='';
              end
              else
                begin
                  val(temp,tmp);
                  if flag then tmp:=-tmp;
                  flag:=false;
                  inc(num,tmp);
                  temp:='';
                end;
          end;
      if left[i]='-' then flag:=true;
      if (i=t)and(left[i] in ['0'..'9']) then
        begin
          val(temp,tmp);
          if flag then tmp:=-tmp;
          flag:=false;
          inc(num,tmp);
          temp:='';
        end;
    end;
  t:=length(right);
  tot:=-tot;
  for i:=1 to t do
    begin
      if right[i] in ['0'..'9'] then temp:=temp+right[i]
        else
          begin
            if right[i]='x' then
              begin
                if temp='' then tmp:=1 else val(temp,tmp);
                if flag then tmp:=-tmp;
                flag:=false;
                inc(tot,tmp);
                temp:='';
              end
              else
                begin
                  val(temp,tmp);
                  if not flag then tmp:=-tmp;
                  inc(num,tmp);
                  flag:=false;
                  temp:='';
                end;
          end;
      if right[i]='-' then flag:=true;
      if (i=t)and(right[i] in ['0'..'9']) then
        begin
          val(temp,tmp);
          if not flag then tmp:=-tmp;
          flag:=false;
          inc(num,tmp);
          temp:='';
        end;
    end;
  if num=0 then ans:=0 else 
  ans:=num/tot;
  writeln(ans:0:3);
  close(input);close(output);
end.

3.MUTI

貌似是数学题，我开始也一直向数学解法上想。

其实是一个经典的完全无限背满包

f[i]=sigma(f[i-a[j]);

 

#include<fstream>
using namespace std;
ifstream cin("muti.in");
ofstream cout("muti.out");
int n,c;
int a[110];
int f[100010];
int main()
{
    cin>>n>>c;
    for (int i=1;i<=n;i++) cin>>a[i];
    f[0]=1;
    for (int i=1;i<=n;i++)
    {
        for (int j=a[i];j<=c;j++)
        {
            f[j]+=f[j-a[i]];
            f[j]%=999983;
        }
    }
    cout<<f[c]<<endl;
    return 0;
}

 

4.交错匹配

DP

f[i][j]=max{f[i][j-1],f[i-1][j]}

找到距离i最近的a[s]和b[k]相等f[i][j]>?=f[s-1][k-1];

#include<fstream>
using namespace std;
ifstream cin("cross.in");
ofstream cout("cross.out");
int m,n;
int a[1010],b[1010],s[32768];
int k,f[1010][1010];
int main(void)
{
    int i,j;
    cin>>n>>m;
    for (i=1;i<=n;i++) cin>>a[i];
    for (i=1;i<=m;i++) cin>>b[i];
    for (i=0;i<=32767;i++) s[i]=10000;
    for (i=1;i<=n;i++)
    {
        k=0;
        for (j=1;j<=m;j++)
        {
            if (a[i]==b[j]) {k=j;}
            if (a[i]!=b[j]) 
              if ((s[b[j]]<i)&&(k>0)) f[i][j]>?=(f[s[b[j]]-1][k-1]+2);
            f[i][j]>?=f[i][j-1];
            f[i][j]>?=f[i-1][j];
        }
        s[a[i]]=i;
    }
    cout<<f[n][m]<<endl;
    return 0;
}