ZOJ-The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple-E

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ACM模版

描述

题解

典型的数位 dp，没什么太多可说的，注意前缀 0 不需要剔除。

代码

#include <cstdio>#include <iostream>using namespace std;typedef long long ll;const int LEN = 8;const int MAXN = LEN + 5;const int EC[] = {6, 2, 5, 5,                  4, 5, 6, 3,                  7, 6, 6, 5,                  4, 5, 5, 4};const ll INF = 0xFFFFFFFF;const ll POW[] = {1, 16, 256, 4096,    65536, 1048576, 16777216, 268435456};char s[MAXN];ll d[MAXN];ll dp[MAXN];  //  前x位中1出现次数（0~9均是如此）void init(){    for (int i = 1; i < MAXN; i++)    {        dp[i] = dp[i - 1] * 16 + POW[i - 1];    }}int HtoD(char ch){    if (ch >= '0' && ch <= '9')    {        return ch - '0';    }    else    {        return ch - 'A' + 10;    }}ll count(int place) //  数是x，查询的数位是place{    ll result = 0;    ll tail = 0;    for (int i = 0; i < LEN; i++)    {        if (d[i] > place)        {            result += POW[i] + d[i] * dp[i];        }        else if (d[i] == place)        {            result += tail + 1 + d[i] * dp[i];        }        else if (d[i] < place)        {            result += d[i] * dp[i];        }        tail = tail + d[i] * POW[i];    //  头为place个数    }    return result;}ll solve(ll x){    if (x < 0)    {        return 0;    }    ll res = 0;    for (int i = 0; i < LEN; i++)    {        d[i] = x % 16;        x /= 16;    }    for (int i = 0; i < 16; i++)    {        res += count(i) * EC[i];    }    return res;}int main(){    init();    int T;    scanf("%d", &T);    int n;    while (T--)    {        scanf("%d%s", &n, s);        ll num = 0;        for (int i = LEN - 1; i >= 0; i--)        {            num += HtoD(s[i]) * POW[LEN - i - 1];        }        ll num_ = num + n - 1, temp = 0;        if (num_ > INF)        {            temp = solve(INF) - solve(num - 1);            num = 0;            num_ -= INF + 1;        }        printf("%lld\n", temp + solve(num_) - solve(num - 1));    }    return 0;}

参考

51Nod 1042 数字0-9的数量

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