題目鏈接

題目描述

AFei loves numbers. He defines the natural number containing “520” as the AFei number, such as 1234520, 8752012 and 5201314. Now he wants to know how many AFei numbers are not greater than n.

輸入描述:

The first line contains an integer T (1 <= T <= 100 ).

The following T lines contain an interger n ( 0 <= n <= 1e18 ).

輸出描述:

For the last T lines, output the total numbers of AFei numbers that are not greater than n.

輸入

2
1000
5520

輸出

1
16

說明

For the first case, only 520 is AFei number.

For the second case, 520,1520, 2520, 3520, 4520, 5200, 5201, 5202, 5203, 5204, 5205, 5206, 5207, 5208, 5209 and 5520 are AFei number. So there are 16 AFei numbers.

思路

計算不包括520的數字

• dp[ i ][ j ] 表示以數字 j 結尾后面剩余 i 位數字中不包括520的數字的數量
• dp[ i ][ j ] = ${\sum }_{k=0}^{k=top}$dp[ i - 1 ][ (j % 10) * 10 + k ]

計算包括520的數字

• dp[ i ][ j ][ flag ] 表示以數字 j 結尾后面剩余 i 位數字中包括520的數字的數量，其中flag表示走到當前位置是否出現過520
• 為什么加上flag表示？ 如果520出現過，遞歸到0的時候就要+1，否者不加。

// 計算不包括 #include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define REP(i, n) for (int i = 1; i <= (n); ++i) #define rep(i, n) for (int i = 0; i < (n); ++i) #define maxn 100005 using namespace std;int num[20]; LL dp[20][100]; LL dfs(int len, int pre, int limit) {if (!len) return 1;// 使用dp數組，前提limit = 0if (!limit && dp[len][pre] != -1) return dp[len][pre];LL sum = 0;int top = 9;if (limit) top = num[len];for (int i = 0; i <= top; ++i) {if (i == 0 && pre == 52) continue;sum += dfs(len-1, (pre%10) * 10 + i, limit&&i==top);}// 只有在limit = 0 時，更新dpif (!limit) dp[len][pre] = sum;return sum; }LL solve(LL n) {int len = 0;LL t = n;while (t) {num[++len] = t % 10;t /= 10;}LL ans = dfs(len, 0, 1);ans = n + 1 - ans;printf("%lld\n", ans);return 0; }int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout); #endifmem(dp, -1);int T;scanf("%d", &T);while (T--) {LL n;scanf("%lld", &n);solve(n);}return 0; } // 計算包括 #include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define REP(i, n) for (int i = 1; i <= (n); ++i) #define rep(i, n) for (int i = 0; i < (n); ++i) #define maxn 100005 using namespace std;int num[20]; LL dp[20][100][2]; LL dfs(int len, int pre, int limit, int flag) {if (!len) return flag ? 1 : 0;if (!limit && dp[len][pre][flag] != -1) return dp[len][pre][flag];int top = limit ? num[len] : 9;LL sum = 0;for (int i = 0; i <= top; ++i) {if (i == 0 && pre == 52) sum += dfs(len-1, (pre%10)*10+i, limit&&i==top, 1);else sum += dfs(len-1, (pre%10)*10+i, limit&&i==top, flag);}if (!limit) dp[len][pre][flag] = sum;return sum; }LL solve(LL n) {int len = 0;LL t = n;while (t) {num[++len] = t % 10;t /= 10;}LL ans = dfs(len, 0, 1, 0);printf("%lld\n", ans);return 0; }int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout); #endifmem(dp, -1);int T;scanf("%d", &T);while (T--) {LL n;scanf("%lld", &n);solve(n);}return 0; }

總結

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