ZOJ3962 计算一段时间数码管所需能量

ZOJ Problem Set - 3962Seven Segment DisplayTime Limit: 2 Seconds      Memory Limit: 65536 KBA seven segment display, or seven segment indicator, is a form of electronic display device for displaying decimal numerals that is an alternative to the more complex dot matrix displays. Seven segment displays are widely used in digital clocks, electronicmeters, basic calculators, and other electronic devices that display numerical information.Edward, a student in Marjar University, is studying the course "Logic and Computer Design Fundamentals" this semester. He bought an eight-digit seven segment display component to make a hexadecimal counter for his course project.In order to display a hexadecimal number, the seven segment display component needs to consume some electrical energy. The total energy cost for display a hexadecimal number on the component is the sum of the energy cost for displaying each digit of thenumber. Edward found the following table on the Internet, which describes the energy cost for display each kind of digit.

Digit Energy Cost(units/s) 0 6 1 2 2 5 3 5 4 4 5 5 6 6 7 3

Digit Energy Cost(units/s) 8 7 9 6 A 6 B 5 C 4 D 5 E 5 F 4

For example, in order to display the hexadecimal number "5A8BEF67" on the component for one second, 5 + 6 + 7 + 5 + 5 + 4 + 6 + 3 = 41 units of energy will be consumed.Edward's hexadecimal counter works as follows:The counter will only work for n seconds. After n seconds the counter will stop displaying.At the beginning of the 1st second, the counter will begin to display a previously configured eight-digit hexadecimal number m.At the end of the i-th second (1 ≤ i < n), the number displayed will be increased by 1. If the number displayed will be larger than the hexadecimal number "FFFFFFFF" after increasing, the counter will set the number to 0and continue displaying.Given n and m, Edward is interested in the total units of energy consumed by the seven segment display component. Can you help him by working out this problem?

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 105), indicating the number of test cases. For each test case:The first and only line contains an integer n (1 ≤ n ≤ 109) and a capitalized eight-digit hexadecimal number m (00000000 ≤ m ≤ FFFFFFFF), their meanings are described above.We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each test case output one line, indicating the total units of energy consumed by the eight-digit seven segment display component.

Sample Input

3
5 89ABCDEF
3 FFFFFFFF
7 00000000

Sample Output

208
124
327

Hint

For the first test case, the counter will display 5 hexadecimal numbers (89ABCDEF, 89ABCDF0, 89ABCDF1, 89ABCDF2, 89ABCDF3) in 5 seconds. The total units of energy cost is (7 + 6 + 6 + 5 + 4 + 5 + 5 + 4) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 6) + (7 + 6 + 6 + 5+ 4 + 5 + 4 + 2) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) = 208.For the second test case, the counter will display 3 hexadecimal numbers (FFFFFFFF, 00000000, 00000001) in 3 seconds. The total units of energy cost is (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 6) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 2) =124.Author: ZHOU, JiayuSource: The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple

dp[i][j]数位 数码管的能量

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <string>#include <cstdio>#include <climits>#include <cmath>#include <vector>#include <set>#include <queue>#include <stack>#include <map>#include <sstream>#define INF 0x3f3f3f3f#define LL long long#define MAXN 100024const long long  F8=4294967295;using namespace std;int power[16] = { 6,2,5,5,4,5,6,3,7,6,6,5,4,5,5,4 };long long s,n,dp[10][100];struct Pair{long long cnt,sum;};int t;int bit[10];long long mmax = 0, ans;long long e16[10];long long dfs(int len,int pre,bool flag){if (len==-1) return pre;if (!flag&&dp[len][pre]!=0) return dp[len][pre];long long ans,tmp;ans=0;int end=flag?bit[len]:15;for (int i=0;i<=end;++i){ans+=dfs(len-1,power[i]+pre,flag&&i==end);}if (!flag) dp[len][pre]=ans;return ans;}long long slove(long long  s){if (s<0) return 0;memset(bit,0,sizeof(bit));int len=0;while (s){bit[len++]=s%16;s/=16;}return dfs(7,0,1);}int main() {e16[0] = 1;for (int i = 1; i<10; ++i) e16[i] = e16[i - 1] * 16;scanf("%d", &t);mmax = 0;while (t--) {ans=0;scanf("%lld%llx", &n, &s);mmax=s+n-1;if (mmax>F8){ans+=slove(F8)-slove(s-1);ans+=slove(mmax-F8-1);}else {ans-=slove(s-1);ans+=slove(mmax);}printf("%lld\n", ans);}return 0;}

一开始写的是用数量*当前位的能量+回溯回来的能量 忘记了pre就是整个数码管的能量 总想着恨7不成妻的与7无关的数的个数 就写成这样了

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <string>#include <cstdio>#include <climits>#include <cmath>#include <vector>#include <set>#include &l9b0dt;queue>#include <stack>#include <map>#include <sstream>#define INF 0x3f3f3f3f#define LL long long#define fora(i,a,n) for(int i=a;i<=n;i++)#define fors(i,n,a) for(int i=n;i>=a;i--)#define sci(x) scanf("%d",&x)#define scl(x) scanf("%lld",&x)#define MAXN 100024const long long  F8=4294967295;using namespace std;int power[16] = { 6,2,5,5,4,5,6,3,7,6,6,5,4,5,5,4 };long long s,n;struct Pair{long long cnt,sum;}dp[10][100];int t;int bit[10];long long mmax = 0, ans;long long e16[10];Pair dfs(int len,int pre,bool flag){if (len==-1){Pair tmp;tmp.sum=0;tmp.cnt=1;return tmp;}if (!flag&&dp[len][pre].cnt!=0) return dp[len][pre];Pair ans,tmp;ans.sum=ans.cnt=0;int end=flag?bit[len]:15;for (int i=0;i<=end;++i){tmp=dfs(len-1,power[i]+pre,flag&&i==end);ans.sum+=tmp.sum+tmp.cnt*power[i];ans.cnt+=tmp.cnt;}if (!flag) dp[len][pre]=ans;return ans;}long long slove(long long  s){if (s<0) return 0;memset(bit,0,sizeof(bit));int len=0;while (s){bit[len++]=s%16;s/=16;}return dfs(7,0,1).sum;}int main() {e16[0] = 1;for (int i = 1; i<10; ++i) e16[i] = e16[i - 1] * 16;scanf("%d", &t);mmax = 0;while (t--) {ans=0;scanf("%lld%llx", &n, &s);mmax=s+n-1;if (mmax>F8){ans+=slove(F8)-slove(s-1);ans+=slove(mmax-F8-1);}else {ans-=slove(s-1);ans+=slove(mmax);}printf("%lld\n", ans);}return 0;}