标签：lin sample find namespace include display answer targe src
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c?>?a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1?≤?a,?b?≤?105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
1 7
7
100 47
147
解题思路：从a开始遍历直到找到c，使c从左读到右的幸运数的组合数等于b，利用string的begin和end函数和reverse函数反转string.
ac代码：
#include<iostream> #include<string> #include<algorithm> using namespace std; string mask(int a){ string ret; while(a){ if(a%10==4||a%10==7) { ret+=((a%10)+‘0‘); } a/=10; } reverse(ret.begin(),ret.end()); return ret; } int main() { int a; string b; cin>>a>> b; int res= a + 1; while(mask(res)!= b){ res++; } cout<<res<<endl; return 0; }
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves long lucky numbers very much. He is interested in the minimum lucky number d that meets some condition. Let cnt(x) be the number of occurrences of number x in number d as a substring. For example, if d?=?747747, then cnt(4)?=?2, cnt(7)?=?4, cnt(47)?=?2, cnt(74)?=?2. Petya wants the following condition to fulfil simultaneously: cnt(4)?=?a1, cnt(7)?=?a2, cnt(47)?=?a3, cnt(74)?=?a4. Petya is not interested in the occurrences of other numbers. Help him cope with this task.
Input
The single line contains four integers a1, a2, a3 and a4 (1?≤?a1,?a2,?a3,?a4?≤?106).
Output
On the single line print without leading zeroes the answer to the problem — the minimum lucky number d such, that cnt(4)?=?a1, cnt(7)?=?a2, cnt(47)?=?a3, cnt(74)?=?a4. If such number does not exist, print the single number "-1" (without the quotes).
Examples
2 2 1 1
4774
4 7 3 1
-1
解题思路：
如果c>d，第一个和最后一个分别是4和7，中间输出d个74，要使构成的数最小，4一定放在前面输出，7放在后面输出
如果c<d，第一个和最后一个分别是7和4，中间输出c个47，4一定在前面输出，7在最后一个4之前输出
如果c=d,如果a>c,则以4为开头和结尾输出，输出a-c-1个4，再输出c个47，b-c个7，再输出一个4
如果a=c，以7为开头和结尾输出，输出c个47，b-c-1个7
ac代码：
#pragma comment(linker,"/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<map> #include<cmath> #include<vector> using namespace std; #define pf printf #define sf scanf #define for1(i,a) for(int (i)=1;(i)<=(a);(i)++) int a,b,c,d; int main() { cin>>a>>b>>c>>d; if(abs(c-d)>1||a+b<=c+d||a<c||b<d||a<d||b<c) { puts("-1"); } else if(c>d) { a=a-d,b=b-d; for1(i,a)pf("4"); for1(i,d)pf("74"); for1(i,b)pf("7"); puts(""); } else if(c==d) { if(a>c) { a-=1+c; b-=c; for1(i,a)pf("4"); for1(i,c)pf("47"); for1(i,b)pf("7"); pf("4"); puts(""); } else { b-=1+c; pf("7"); for1(i,c)pf("47"); for1(i,b)pf("7"); puts(""); } } else { a-=c+1,b-=c+1; pf("7"); for1(i,a)pf("4"); for1(i,c)pf("47"); for1(i,b)pf("7"); pf("4"); puts(""); } return 0; }
Codeforces Round #104 (Div.2)补题报告
标签：lin sample find namespace include display answer targe src
原文地址：https://www.cnblogs.com/nanan/p/14150886.html
