5184. 【NOIP2017提高组模拟6.29】Gift (Standard IO)

Description

 

Input

 

Solution

设dp(i,j)表示选取1-i个物品，剩余为j的方案数。

dp(n+1,m)=1，dp(i,j)=dp(i+1,j)+(i+1,j+a[i])

那么如何统计最终答案呢？

经过上面的讨论，我们可以知道：

设前缀和 sum[i] 表示 从 a[1] 到 a[i] 的和，则对于一个 i 有：

0≤m−sum[i]−j<a[i]

解出花费 j 循环统计即可。

 

代码

const
  mo=10000007;
var
  n,m,ans:longint;
  a,sum:array [0..1001] of longint;
  f:array [0..1001,0..1001] of longint;
function min(o,p:longint):longint;
begin
  if o<p then exit(o);
  exit(p);
end;

procedure init;
var
  i:longint;
begin
  readln(n,m);
  for i:=1 to n do
    read(a[i]);
end;

procedure qsort(l,r:longint);
var
  i,j,mid,t:longint;
begin
  if l>r then exit;
  i:=l; j:=r;
  mid:=a[(l+r) div 2];
  repeat
    while a[i]<mid do inc(i);
    while a[j]>mid do dec(j);
    if i<=j then
      begin
        t:=a[i]; a[i]:=a[j]; a[j]:=t;
        inc(i); dec(j);
      end;
  until i>j;
  qsort(l,j);
  qsort(i,r);
end;

procedure main;
var
  i,j,k:longint;
begin
  f[n+1,m]:=1;
  for i:=1 to n do
    sum[i]:=sum[i-1]+a[i];
  if sum[n]<=m then
    begin
      ans:=1;
      exit;
    end;
  ans:=0;
  for i:=n downto 1 do
    begin
      j:=sum[i-1];
      k:=min(m+1,sum[i-1]+a[i]);
      while j<k do
        begin
          ans:=(ans+f[i+1,j]) mod mo;
          inc(j);
        end;
      for j:=m downto 0 do
        begin
          f[i,j]:=(f[i,j]+f[i+1,j]) mod mo;
          if j+a[i]<=m then
            f[i,j]:=(f[i,j]+f[i+1,j+a[i]]) mod mo;
        end;
    end;
end;

begin
  init;
  qsort(1,n);
  main;
  writeln(ans);
end.