HUD 2089
不要62
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 43142 Accepted Submission(s): 15815
Problem Description
杭州人称那些傻乎乎粘嗒嗒的人为62（音：laoer）。
杭州交通管理局经常会扩充一些的士车牌照，新近出来一个好消息，以后上牌照，不再含有不吉利的数字了，这样一来，就可以消除个别的士司机和乘客的心理障碍，更安全地服务大众。不吉利的数字为所有含有4或62的号码。例如：62315 73418 88914都属于不吉利号码。但是，61152虽然含有6和2，但不是62连号，所以不属于不吉利数字之列。你的任务是，对于每次给出的一个牌照区间号，推断出交管局今次又要实际上给多少辆新的士车上牌照了。Input
输入的都是整数对n、m（0<n≤m<1000000），如果遇到都是0的整数对，则输入结束。
Output
对于每个整数对，输出一个不含有不吉利数字的统计个数，该数值占一行位置。
Sample Input
1 100 0 0
Sample Output
80
Author
qianneng
做法一：
#include#include #include using namespace std;int dp[10][3];void init(){ int len = 6; memset(dp, 0, sizeof(dp)); dp[0][0] = 1; for(int i = 1; i <= len; i++) { dp[i][0] = dp[i-1][0] * 9 - dp[i-1][1]; //首位不放4 不在次位是2的情况下放6 dp[i][1] = dp[i-1][0];//首位放的是2 dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1] + dp[i-1][0]; //在2前面放6、放4 或者 在不合格的情况下随便放 }}int solve(int x){ int a[10],n = x, ans = 0; int len = 0; int flag = 0; while (x) { a[++len] = x%10; x /= 10; } a[len + 1] = 0; for(int i = len; i >= 1; i--) { ans += dp[i-1][2] * a[i]; if(flag) { ans += dp[i-1][0] * a[i]; } if(!flag && a[i] > 4) { ans += dp[i-1][0]; } if(!flag && a[i+1] == 6 && a[i] > 2) { ans += dp[i][1]; } if(!flag && a[i] > 6) { ans += dp[i-1][1]; } if(a[i] == 4 || (a[i] == 2 && a[i+1] == 6)) { flag = 1; } } return n - ans;}int main(){ int l, r; init(); while(scanf("%d%d", &l, &r), l || r) { printf("%d\n", solve(r + 1) - solve(l)); } return 0;}
做法二（记忆化搜索）：
#include#include #include using namespace std;int a[10];int dp[10][2];int dfs(int pre, int pos, int status, bool limit){ if(pos == 0) return 1; if(!limit && dp[pos][status] != -1) return dp[pos][status]; int up = limit ? a[pos] : 9; int ans = 0; for(int i = 0; i <= up; i++) { if(i == 4 ) continue; if(pre == 6 && i == 2) continue; ans += dfs(i, pos-1, i == 6, limit && i == up); } if(!limit) dp[pos][status] = ans; return ans;}int solve(int x){ int len = 0; memset(dp, -1, sizeof(dp)); while(x) { a[++len] = x%10; x /= 10; } return dfs(-1, len, 0, true);}int main(){ int l, r; while(scanf("%d%d", &l, &r), l || r) { printf("%d\n", solve(r) - solve(l - 1)); } return 0;}
有了上面的思路，可以解决其他类似的问题。
HDU 3555 试试水
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 18374 Accepted Submission(s): 6781
Problem Description
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?Input
The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.
The input terminates by end of file marker.Output
For each test case, output an integer indicating the final points of the power.
Sample Input
3 1 50 500
Sample Output
0 1 15
Hint
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.
Author
做法一：
#include#include #include using namespace std;typedef long long LL;LL dp[30][3];int a[30];void init(){ memset(dp, 0, sizeof(dp)); dp[0][0] = 1; for(int i = 1; i <= 25; i++) { dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1]; //减去首位是6 次位是2 dp[i][1] = dp[i-1][0]; //首位是2 dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1];//加上首位是6 次位是2 }}LL solve(LL x){ int len = 0; while(x) { a[++len] = x%10; x /= 10; } a[len + 1] = 0; LL ans = 0LL; bool flag = false; for(int i = len; i >= 1; i--) { ans += dp[i-1][2] * a[i]; if(flag) ans += dp[i-1][0] * a[i]; if(!flag && a[i] > 4) ans += dp[i-1][1]; if(a[i+1] == 4 && a[i] == 9) flag = true; } return ans;}int main(){ init(); int q; LL n; scanf("%d", &q); while(q--) { scanf("%lld\n", &n); printf("%lld\n", solve(n+1)); } return 0;}
做法二（记忆化搜索）：
#include#include #include typedef long long LL;int a[30];LL dp[30][3];LL dfs(int pre, int pos, int status, bool limit){ if(pos == 0) return status == 2; if(!limit && dp[pos][status] != -1) { return dp[pos][status]; } LL ans = 0LL; int up = limit ? a[pos] : 9; for(int i = 0; i <= up; i++) { int nstatus; if(status == 2 || pre == 4 && i == 9) nstatus = 2; else if(i == 4) nstatus = 1; else nstatus = 0; ans += dfs(i, pos-1, nstatus, limit && i == up); } if(!limit) dp[pos][status] = ans; return ans;}LL solve(LL x){ LL n = x, len = 0; memset(dp, -1, sizeof(dp)); while(x) { a[++len] = x%10; x /= 10; } a[len + 1] = 0; return dfs(-1, len, 0, true);}int main(){ LL n; int q; scanf("%d", &q); while(q--) { scanf("%lld", &n); printf("%lld\n",solve(n)); } return 0;}