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
3150500
Sample Output
0115
题意：计算一个数内含连续数49的个数
思路
：和上一篇博客题意相近，，同理可解，数据很大，要用__int64
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i < b; i++)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d%d", &a, &b)#define sfff(a,b,c) scanf("%d%d%d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 22ll dp[N][3];//dp[i][0] 没有不吉利//dp[i][1] 没有不吉利 首位 为 9//dp[i][2] 有不吉利void inint(){dp[0][0]=1;ll i;for(i=1;i<N;i++){dp[i][0]=dp[i-1][0]*10-dp[i-1][1];dp[i][1]=dp[i-1][0];dp[i][2]=dp[i-1][2]*10+dp[i-1][1];}}void solve(ll x){ll i,len=0,bit[N];while(x){bit[++len]=x%10;x=x/10;}bit[len+1]=0;ll ans=0;bool flag=false;for(i=len;i>=1;i–){ans+=bit[i]*dp[i-1][2];if(flag)ans+=bit[i]*dp[i-1][0];if(flag) continue;if(bit[i]>4)ans+=dp[i-1][1];if(bit[i+1]==4&&bit[i]==9)flag=true;}printf("%I64d\n",ans);}int main(){int t;ll le;inint();sf(t);while(t–){scanf("%I64d",&le) ;solve(le+1);}return 0;}
带着我的相机和电脑，远离繁华，走向空旷。
