Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).
Input
The first line has a number T (T <= 10000) , indicating the number of test cases. For each test case, there are two numbers A and B (0 <= A,B < 10[sup]9[/sup])
Output
For every case,you should output "Case #t: " at first, without quotes. The [I]t[/I] is the case number starting from 1. Then output the answer.
Sample Input
3
0 100
1 10
5 100
Sample Output
Case #1: 1
Case #2: 2
Case #3: 13
其实转化后的数字比原来的要小得多 一开始还纠结开不起数组
把数位的和保存起来 最后读取完的时候再比较即可
为了memset优化 dp数组用减法
#include<bits/stdc++.h> using namespace std; //input by bxd #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define repp(i,a,b) for(int i=(a);i>=(b);--i) #define RI(n) scanf("%d",&(n)) #define RII(n,m) scanf("%d%d",&n,&m) #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k) #define RS(s) scanf("%s",s); #define ll long long #define REP(i,N) for(int i=0;i<(N);i++) #define CLR(A,v) memset(A,v,sizeof A) ////////////////////////////////// #define inf 0x3f3f3f3f #define N 20 int f(int x) { if(!x)return 0; int ans=f(x/10); return ans*2+(x%10); } ll dp[20][10000+5]; ll a[N]; int all; ll dfs(int pos,int sum,bool lead,bool limit) { if(!pos) { return sum<=all; } if(sum>all)return 0; if(!limit&&!lead&&dp[pos][all-sum]!=-1)return dp[pos][all-sum]; ll ans=0; int up=limit?a[pos]:9; rep(i,0,up) { ans+=dfs(pos-1, sum+i*(1<<pos-1) , lead&&i==0,limit&&i==a[pos]); } if(!limit&&!lead)dp[pos][all-sum]=ans; return ans; } ll solve(int b) { int pos=0; while(b) { a[++pos]=b%10; b/=10; } return dfs(pos, 0 ,true,true); } int main() { CLR(dp,-1); RI(cas); int kase=0; while(cas--) { int a,b; cin>>a>>b; all=f(a); printf("Case #%d: %lld ",++kase,solve(b)); } return 0; }