LightOJ - 1140 统计0的数位 数位DP

注意以下几点：

搜索维度非约束条件的都要记录，否则大概率出错，比如_0

st参数传递和_0的互相影响要分辨清楚

num==-1就要返回0而不是1

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 30;

const int MOD = 2520;

const double eps = 1e-10;

typedef long long ll;

const int oo = 0x3f3f3f3f;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

ll a[maxn],dp[maxn][maxn];

ll DP(int cur,int limit,int _0,int st){

    if(cur==0&&st) return 1;

    if(cur==0) return _0;

    if(dp[cur][_0]!=-1&&!limit&&!st) return dp[cur][_0];

    int up=limit?a[cur]:9;

    ll ans=0;

    rep(i,0,up){

        ans+=DP(cur-1,limit&&a[cur]==i,(st&&i==0)?0:_0+(i==0),st&&(i==0));

    }

    return (limit||st)?ans:dp[cur][_0]=ans;

}

ll solve(ll num){

    if(num==-1)return 0;

    if(num==0) return 1;

    int cur=0;

    while(num){

        a[++cur]=num%10;

        num/=10;

    }

    return DP(cur,1,0,1);

}

int main(){

    memset(dp,-1,sizeof dp);

    int T=read(),kase=0;

    while(T--){

        ll l=read();

        ll r=read();

        printf("Case %d: ",++kase);

        println(solve(r)-solve(l-1));

    }

    return 0;

}