Đề bài:
Thuật toán:
- Bài này sử dụng phương pháp đệ quy có nhớ
- (thuật toán đang cập nhập)
Bạn có thể đọc code để hiểu rõ hơn.
Code:
Pascal
const
fi='';
fo='';
maxn=20;
var
a,b,tg1,tg2 : int64;
aa : array[1..maxn] of longint;
i,j,n,k,d : longint;
f : array[0..maxn,0..maxn,0..maxn,false..true,false..true] of int64;
procedure swap(var x,y : longint);
var tg : longint;
begin
tg:=x;x:=y;y:=tg;
end;
procedure chuanbi(x : int64; var n : longint);
var i,j : longint;
begin
n := 0;
while x>0 do
begin
inc(n); aa[n] := x mod 10;
x := x div 10;
end;
i:=1; j:=n;
while (i -1 then exit(f[i,tr,sl,big0,small]);
if i>n then
if (sl<=k) and (small) then exit(1) else exit(0);
dem := 0;
if not small then
begin
for j := 0 to aa[i] do
begin
if big0 then tg := sl + ord(abs(j-tr)<=d) else tg := 0;
dem := dem + tinh(i+1,j,tg,big0 or (j>0),small or (j0),small or (j
C++
#include
#define FORE(i, a, b) for(int i = a; i <= b; i++)
#define FORD(i, a, b) for(int i = a; i >= b; i--)
#define FOR(i, a, b) for(int i = a; i < b; i++)
const int MAXN = 1e5 * 5;
const int INF = 1e9 + 7;
using namespace std;
long long l, r, D, K, n;
long long f[20][10][20][2][2];
int a[20];
bool check(long long x)
{
bool ans = 1;
int top = 0;
int tmp = x;
while (tmp){
a[++top] = tmp % 10;
tmp /= 10;
}
reverse(a + 1, a + top + 1);
int dem = 0;
FORE(i, 2, top) if (abs(a[i] - a[i - 1]) <= D) dem++;
//if (x == 103) cout << dem<<" "< n){
return (worse <= K && greater0 > 0);
}
if (f[i][dig][worse][ok][greater0] > -1) return f[i][dig][worse][ok][greater0];
long long ans = 0;
if (i == 1){
FORE(x, 0, a[i]) ans += duyet(i + 1, x, worse, ok | (x < a[i]), (x > 0));
} else {
int last = 9;
if (ok == 0) last = a[i];
int wnext;
//if (i == 3) cout << last << "??"< 0));
}
}
f[i][dig][worse][ok][greater0] = ans;
return ans;
}
int main()
{
ios::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("DEMSO.inp", "r", stdin);
freopen("DEMSO.out", "w", stdout);
#endif //MIKELHPDATKE
cin >> l >> r >> D >> K;
//trau();
long long tmp = r, top = 0;
while (tmp){
a[++top] = tmp % 10;
tmp /= 10;
}
reverse(a + 1, a + top + 1); n = top;
memset(f, -1, sizeof(f));
long long ans = duyet(1, 0, 0, 0, 0);
// cout << ans << endl;
tmp = l - 1, top = 0;
while (tmp){
a[++top] = tmp % 10;
tmp /= 10;
}
reverse(a + 1, a + top + 1); n = top;
memset(f, -1, sizeof(f));
if (l - 1 == 0) cout << ans;
else{
ans -= duyet(1, 0, 0, 0, 0);
cout << ans;
}
//cout<
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