题面

题目链接

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备用题面

将 $N$ 到 $M$ 的数字依次写在一张纸上，问出现多少个 $0$ 。

思路

就是一道数位DP板子题，不想说话了。

代码

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// #pragma GCC optimize(2)
// #pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastIO 
{
    // char buf[1 << 20], *p1, *p2;
    // #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
};
using namespace FastIO;

ll n, m;

int a[110];
ll dp[110][110];

inline void Input() {
    read(n, m);
}

inline ll Dfs(int len, int lim, int top, int sum) {
    if(len == 0) {
        if(top) return 1;
        return sum;
    }
    if(!lim && !top && dp[len][sum] != -1) return dp[len][sum];
    int mx = lim ? a[len] : 9;
    ll ans = 0;
    for(int i = 0; i <= mx; i++) {
        ans += Dfs(
            len - 1, 
            lim && i == mx, 
            top && i == 0, 
            top ? 0 : sum + (i == 0)
        );
    }
    if(!lim && !top) dp[len][sum] = ans;
    return ans;
}

inline ll Calc(ll x) {
    if(x < 0) return 0;
    int len = 0;
    while(x) {
        a[++len] = x % 10;
        x /= 10;
    }
    return Dfs(len, 1, 1, 0);
}

inline void Work() {
    memset(dp, -1, sizeof(dp));
    printf("%lld", Calc(m) - Calc(n - 1));
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}