BNUOJ 52325 Increasing or Decreasing 数位dp

题目链接：

https://acm.bnu.edu.cn/v3/contest_show.php?cid=8506#problem/I

I. Increasing or Decreasing

Case Time Limit: 1000ms
Memory Limit: 524288KB

题意

求[l,r]上，数位满足非递增获非递减的数的个数。

题解

1、dp[i][j][k]表示低i位的第i位为j的，k1时表示满足非递减，k2时满足非递增，k==0时表示每一位都相等。

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=22;

int arr[maxn],tot;
///type根据dp具体维数调整
LL dp[maxn][11][4];
///ismax标记表示前驱是否是边界值
///ser标记前驱是否是前导零
LL dfs(int len,int dig,int type,bool ismax,bool iszer) {
if (len == 0) {
///递归边界，这说明前驱都合法了
return 1LL;
}
if (!ismax&&dp[len][dig][type]>=0) return dp[len][dig][type];
LL res = 0;
int ed = ismax ? arr[len] : 9;

///这里插入递推公式
for (int i = 0; i <= ed; i++) {
if(i0&&iszer){
///处理前导零
res+=dfs(len-1,10,0,ismax&&ied,1);
}else{
if(type0){
if(idig||dig10) res+=dfs(len-1,i,0,ismax&&ied,0);
else if(i>dig) res+=dfs(len-1,i,1,ismax&&ied,0);
else if(i<dig) res+=dfs(len-1,i,2,ismax&&ied,0);
}else if(type1){
if(i>=dig){
res+=dfs(len-1,i,type,ismax&&ied,0);
}
}else if(type2){
if(i<=dig){
res+=dfs(len-1,i,type,ismax&&ied,0);
}
}
}
}
return ismax ? res : dp[len][dig][type] = res;
}

LL solve(LL x) {
tot = 0;
while (x) { arr[++tot] = x % 10; x /= 10; }
LL ret=0;
return dfs(tot,10,0,true,true);
}

void init() {
clr(dp,-1);
}

int main() {
init();
int n; scf("%d",&n);
while(n--){
LL l,r;
scf("%lld%lld",&l,&r);
prf("%lld\n",solve(r)-solve(l-1));
}
return 0;
}

//end-----------------------------------------------------------------------

2、状态和上面差不多，不过是固定k，分别求出来，既ans=非递增+非递减-每个数位都相同。

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=22;

int arr[maxn],tot;
///type根据dp具体维数调整
LL dp[maxn][11][4];
///ismax标记表示前驱是否是边界值
///ser标记前驱是否是前导零
LL dfs(int len,int dig,int type,bool ismax,bool iszer) {
if (len == 0) {
///递归边界，这说明前驱都合法了
return 1LL;
}
if (!ismax&&dp[len][dig][type]>=0) return dp[len][dig][type];
LL res = 0;
int ed = ismax ? arr[len] : 9;

///这里插入递推公式
for (int i = 0; i <= ed; i++) {
if(i0&&iszer){
///处理前导零
res+=dfs(len-1,10,type,ismax&&ied,true);
}else{
if(type1){
if(i>=dig||dig10) res+=dfs(len-1,i,type,ismax&&ied,false);
}else if(type2){
if(i<=dig||dig10) res+=dfs(len-1,i,type,ismax&&ied,false);
}else if(type0){
if(idig||dig10) res+=dfs(len-1,i,type,ismax&&ied,false);
}
}
}
return ismax ? res : dp[len][dig][type] = res;
}

LL solve(LL x) {
tot = 0;
while (x) { arr[++tot] = x % 10; x /= 10; }
LL ret=0;
ret+=dfs(tot,10,1,true,true);
ret+=dfs(tot,10,2,true,true);
ret-=dfs(tot,10,0,true,true);
return ret;
}

void init() {
clr(dp,-1);
}

int main() {
init();
int n; scf("%d",&n);
while(n--){
LL l,r;
scf("%lld%lld",&l,&r);
prf("%lld\n",solve(r)-solve(l-1));
}
return 0;
}

//end-----------------------------------------------------------------------