标签：cal ons climits ace discuss cli big fun tput
As we all know,a palindrome number is the number which reads the same backward as forward,such as 666 or 747.Some numbers are not the palindrome numbers in decimal form,but in other base,they may become the palindrome number.Like 288,it’s not a palindrome number under 10-base.But if we convert it to 17-base number,it’s GG,which becomes a palindrome number.So we define an interesting function f(n,k) as follow:
f(n,k)=k if n is a palindrome number under k-base.
Otherwise f(n,k)=1.
Now given you 4 integers L,R,l,r,you need to caluclate the mathematics expression ∑Ri=L∑rj=lf(i,j)∑i=LR∑j=lrf(i,j) .
When representing the k-base(k>10) number,we need to use A to represent 10,B to represent 11,C to repesent 12 and so on.The biggest number is Z(35),so we only discuss about the situation at most 36-base number.
Input
The first line consists of an integer T,which denotes the number of test cases.
In the following T lines,each line consists of 4 integers L,R,l,r.
(1≤T≤105,1≤L≤R≤109,2≤l≤r≤361≤T≤105,1≤L≤R≤109,2≤l≤r≤36)
Output
For each test case, output the answer in the form of “Case #i: ans” in a seperate line.
Sample Input
3
1 1 2 36
1 982180 10 10
496690841 524639270 5 20
Sample Output
Case #1: 665
Case #2: 1000000
Case #3: 447525746
分析：很好的数位dp的练习；
dp[pos][len][base]表示当前在pos位置，len长度，base进制下的状态；
dfs过程时只需pos小于长度一半时check一下即可；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #include <cassert> #include <ctime> #define rep(i,m,n) for(i=m;i<=(int)n;i++) #define inf 0x3f3f3f3f #define mod 1000000007 #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") #define ls (rt<<1) #define rs (rt<<1|1) #define all(x) x.begin(),x.end() const int maxn=1e5+10; const int N=1e4+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qmul(ll p,ll q,ll mo){ll f=0;while(q){if(q&1)f=(f+p)%mo;p=(p+p)%mo;q>>=1;}return f;} ll qpow(ll p,ll q,ll mo){ll f=1;while(q){if(q&1)f=f*p%mo;p=p*p%mo;q>>=1;}return f;} int n,m,k,t,num[40],tmp[40],cas; ll dp[40][40][40]; ll dfs(int pos,int len,int bs,int zero,int lim) { if(pos<0)return zero; if(zero&&lim&&dp[pos][len][bs]!=-1)return dp[pos][len][bs]; ll ret=0; int ma=lim?bs-1:num[pos],i; rep(i,0,ma) { if(!zero&&i==0)ret+=dfs(pos-1,len-1,bs,zero,lim||i<num[pos]); else if(pos>=(len+1)/2) { tmp[pos]=i; ret+=dfs(pos-1,len,bs,zero||i,lim||i<num[pos]); } else if(pos<(len+1)/2&&i==tmp[len-pos]) { ret+=dfs(pos-1,len,bs,zero||i,lim||i<num[pos]); } } return zero&&lim?dp[pos][len][bs]=ret:ret; } ll gao(ll x,int l,int r) { ll ret=0; ll y=x; int i; rep(i,l,r) { x=y; int pos=0; while(x)num[pos++]=x%i,x/=i; ll p=dfs(pos-1,pos-1,i,0,0); ret+=p*i+y-p; } return ret; } int main(){ int i,j; memset(dp,-1,sizeof(dp)); scanf("%d",&t); while(t--) { ll L,R,l,r; scanf("%lld%lld%lld%lld",&L,&R,&l,&r); --L; printf("Case #%d: %lld\n",++cas,gao(R,l,r)-gao(L,l,r)); } return 0; }
标签：cal ons climits ace discuss cli big fun tput
原文地址：http://www.cnblogs.com/dyzll/p/7427823.html
