AcWing - 338 - 计数问题 = 数位dp

https://www.acwing.com/problem/content/description/340/

第一次做这种数数字的个数的，感觉理论上是差不多的，返回的不是1而是他的贡献罢了。

按道理要注意0的，但是题目里没有0。0毕竟是很特殊的，他全是前导0但也会贡献1个0。

#include
using namespace std;
typedef long long ll;

int a[40];
ll dp[10][40][40];

ll dfs(int id, int pos, int s1, bool lead, bool limit) {
    if(pos == -1) {
        return s1;
    }
    if(!limit && !lead && dp[id][pos][s1] != -1)
        return dp[id][pos][s1];
    int up = limit ? a[pos] : 9;
    ll ans = 0;
    for(int i = 0; i <= up; i++) {
        int ns1 = s1 + (i == id);
        if(id == 0)
            ns1 = s1 + ((i == id) && (!lead));
        ans += dfs(id, pos - 1, ns1, lead && i == 0, limit && i == a[pos]);
    }
    if(!limit && !lead)
        dp[id][pos][s1] = ans;
    return ans;
}

ll ans[10];

void solve(ll x) {
    memset(ans, 0, sizeof(ans));
    if(x <= 0)
        return;
    int pos = 0;
    while(x) {
        a[pos++] = x % 10;
        x /= 10;
    }
    for(int id = 0; id <= 9; ++id )
        ans[id] = dfs(id, pos - 1, 0, true, true);
}

void solve2(ll x) {
    if(x <= 0)
        return;
    int pos = 0;
    while(x) {
        a[pos++] = x % 10;
        x /= 10;
    }
    for(int id = 0; id <= 9; ++id )
        ans[id] -= dfs(id, pos - 1, 0, true, true);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif
    memset(dp, -1, sizeof(dp));

    ll le, ri;
    while(~scanf("%lld%lld", &le, &ri)) {
        if(le == 0 && ri == 0)
            break;
        if(le > ri)
            swap(le, ri);
        solve(ri);
        solve2(le - 1);
        for(int id = 0; id <= 9; ++id) {
            printf("%lld%c", ans[id], " \n"[id == 9]);
        }
    }
}

转载于:https://www.cnblogs.com/Inko/p/11603171.html