poj 2282 The Counting Problem && poj 3286 How many 0's? （数位dp）

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两个题几乎差不多，就是poj2282多求下1-9的数量，求1-9的数量和51nod的1009一样点击打开链接。

数0的情况就有点不一样了，因为前导0不算在内，所以递归时要多加个判断是否是最前面的有效数字.

还有就是0（数字0）单独计算。。递归中不考虑0，最后输出的时候判断下如果L是0的话加个1

poj2282代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 20;int dp[maxn][maxn][maxn], a[maxn];int dfs0(int pos, int num, int lead, int limit){    if(pos == -1) return num;    if(!limit && lead && dp[0][pos][num] != -1) return dp[0][pos][num];    int up = limit ? a[pos] : 9;    int tmp = 0;    for(int i = 0; i <= up; i++)    {        if(i)            tmp += dfs0(pos-1, num, 1, limit && a[pos] == i);        else            tmp += dfs0(pos-1, num+lead, lead, limit && a[pos] == i);    }    if(!limit && lead) dp[0][pos][num] = tmp;    return tmp;}int dfs(int pos, int val, int num, int limit){    if(pos == -1) return num;    if(!limit && dp[val][pos][num] != -1) return dp[val][pos][num];    int up = limit ? a[pos] : 9;    int tmp = 0;    for(int i = 0; i <= up; i++)        tmp += dfs(pos-1, val, num+(i==val), limit && a[pos] == i);    if(!limit) dp[val][pos][num] = tmp;    return tmp;}int solve(int x, int val){    int pos = 0;    while(x)    {        a[pos++] = x%10;        x /= 10;    }    if(!val) return dfs0(pos-1, 0, 0, 1);    else return dfs(pos-1, val, 0, 1);}int main(void){    int a, b;    memset(dp, -1, sizeof(dp));    while(cin >> a >> b, a+b)    {        if(a > b) swap(a, b);        for(int i = 0; i <= 9; i++)            printf("%d%c", solve(b, i)-solve(a-1, i), i==9 ? '\n' : ' ');    }    return 0;}

poj3286代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;typedef long long ll;const int maxn = 20;ll dp[maxn][maxn], a[maxn];ll dfs(int pos, int num, int lead, int limit){    if(pos == -1) return num;    if(!limit && lead && dp[pos][num] != -1) return dp[pos][num];    int up = limit ? a[pos] : 9;    ll tmp = 0;    for(int i = 0; i <= up; i++)    {        if(i)            tmp += dfs(pos-1, num, 1, limit && i == a[pos]);        else            tmp += dfs(pos-1, num+lead, lead, limit && i == a[pos]);    }    if(!limit && lead) dp[pos][num] = tmp;    return tmp;}ll solve(ll x){    int pos = 0;    while(x)    {        a[pos++] = x%10;        x /= 10;    }    return dfs(pos-1, 0, 0, 1);}int main(void){    ll l, r;    memset(dp, -1, sizeof(dp));    while(cin >> l >> r, l+r+2)    {        if(!l)            printf("%lld\n", solve(r)+1);        else            printf("%lld\n", solve(r)-solve(l-1));    }    return 0;}

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