hdu 5936 Difference 二分

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题意：挺明显的题意，就是问满足那个等式，有多少y满足。

做法：我学二分的时候在想，左右分一半处理也应该叫二分。这题也就是左右分一半处理，如果我ans[k][i]，代表在k的幂次下，i的各个位上的数/sum_{各个位x} x^k。

不管这个数有多大，那么我分一半来算，最后得到的sum肯定是没影响的，那么我左边算一个sum，右边算一个sum。x=sum-i*100000+sum-i，

我把xx2[i]排个序，二分找一下满足sum-i有多少个就好了。统计答案即可，刺激！

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e5+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;LL ans1[10][maxn],ans2[10][maxn];LL xx1[maxn],xx2[maxn];LL _pow(LL x,int k){    LL a=1;    W(k--)    {        a*=x;    }    return a;}void init(){    FOR(1,9,k)    {        FOr(0,maxn,i)        {            int cnt=i,t=0;            LL ans=0;            W(cnt)            {                int x=cnt%10;                ans+=_pow(x,k);                cnt/=10;            }            ans1[k][i]=ans-(LL)i*100000;            ans2[k][i]=ans-i;        }    }}LL check(LL x){    LL xx=upper_bound(xx2,xx2+100000,x)-lower_bound(xx2,xx2+100000,x);    return xx;}LL x,k;void solve(){    S_2(x,k);    FOR(0,100000,i)    {        xx1[i]=ans1[k][i];        xx2[i]=ans2[k][i];    }    LL cnt=0;    if(x==0) cnt--;    sort(xx2,xx2+100000);    for(int i=0;i<100000;i++)    {        LL z=x-xx1[i];        cnt+=check(z);    }    print(cnt);}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    init();    s_1(t);    for(int cas=1;cas<=t;cas++)    {        printf("Case #%d: ",cas);        solve();    }}