• 有趣的数 zoj 月赛


    题目描述

    让我们来考虑1到N的正整数集合。让我们把集合中的元素按照字典序排列,例如当N=11时,其顺序应该为:1,10,11,2,3,4,5,6,7,8,9。

    定义K在N个数中的位置为Q(N,K),例如Q(11,2)=4。现在给出整数K和M,要求找到最小的N,使得Q(N,K)=M。

    输入输出格式

    输入格式:

    输入文件只有一行,是两个整数K和M。

    输出格式:

    输出文件只有一行,是最小的N,如果不存在这样的N就输出0。

    输入输出样例

    输入样例#1: 复制
    2 4
    
    输出样例#1: 复制
    11
    
    输入样例#2: 复制
    100000001 1000000000
    输出样例#2: 复制
    100000000888888879

    说明

    【数据约定】

    40%的数据,1<=K,M<=10^5;

    100%的数据,1<=K,M<=10^9。

    很像数位dp是吧;

    不得不说还是我太菜了;

    其实直接暴力算即可;

    以233为例,

    我们可以先计算出它最小应该在什么位置,然后与M进行比较,

    如果M还有剩余,那么扩展位数慢慢加上,由于是*10的递增,复杂度是log的;

    可以跑的很快;

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 1000005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9 + 7;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-4
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    inline ll rd() {
    	ll x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    int T;
    int K, M;
    ll maxx[21];
    int dig[2100];
    int a[2000];
    ll qpow(int x) {
    	ll ans = 1;
    	for (int i = 1; i <= x; i++)ans *= 10;
    	return ans;
    }
    
    int main() {
    	//ios::sync_with_stdio(0);
    	cin >> T;
    	maxx[0] = 1;
    	for (int i = 1; i <= 20; i++)maxx[i] = maxx[i - 1] * 10;
    	while (T--) {
    		rdint(K); rdint(M);
    		bool fg = 0;
    		for (int i = 0; i < 19; i++) {
    			if (K == maxx[i] && M != i + 1) {
    				cout << 0 << endl; 
    				fg = 1; break;
    			}
    		}
    		if (fg)continue;
    		ll tp = K; int tot = 0; ms(dig);
    		while (tp) {
    			int y = tp % 10;
    			dig[++tot] = y; tp /= 10;
    		}
    		for (int i = 1; i <= tot; i++) {
    			a[i] = dig[tot - i + 1];
    		}
    		tp = 1; ll reg = 0; ll tmp = 0;
    		for (int i = 1; i <= tot; i++) {
    			reg = reg * 10ll + a[i];
    			tmp += (reg - qpow(i - 1) + 1);
    		}
    		if (tmp > M) {
    			cout << 0 << endl; continue;
    		}
    		if (tmp == M) { cout << K << endl; continue; }
    		tp = 1; M -= tmp;
    		ll number = 1ll * reg * 10 - qpow(tot + tp - 1);
    		while (number < M) {
    			M -= number; number *= 10; tp++;
    		}
    		cout << qpow(tp + tot - 1) + M - 1 << endl;
    	}
    	return 0;
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10293033.html
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