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Given an integer YY, you need to find the minimal integer KK so that there exists a XX satisfying X−Y=Z(Z≥0)X−Y=Z(Z≥0) and the number of different digit between XX and ZZis KK under decimal system.
For example: Y=1Y=1, you can find a X=100X=100 so that Z=99Z=99 and KK is 33 due to 1≠01≠0 and 0≠90≠9. But for minimization, we should let X=1X=1 so that Z=0Z=0 and KK can just be 11.
Only one integer Y(0≤Y≤1018).Y(0≤Y≤1018).
The minimal KK.
| Sample Input | Sample Output |
|---|---|
1 |
1 |
191 |
题解：
X−Y=Z(Z≥0),已知Y，求最小K，K定义为Z与X不相同的位置的个数；例如280 - 191 = 89;
X - Z = Y;
根据减法运算我们可以想到Xi - Zi = Yi; 如果数字相同我们可以得到 Xi = Zi;
即 Xi - Xi = Yi; 所以Yi 为0的位置肯定能找到满足的；再者有可能进位，即：10 +Xi -Xi -1 = 9;
所以9也能得到，但是这两者相互影响，取09相邻只能取一个，还有特殊情况9在最后一位，最高的退位，90000对0的影响。。。考虑清就好了；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; int a[25]; int ans; void dfs(int cur, int cnt, int tp, int kg){ // printf("%d %d\n", cur, tp); if(cur >= tp){ ans = max(ans, cnt); return; } if(a[cur] == 0){ if(kg != 9 && !(cur == tp - 2 && a[tp - 1] == 9))dfs(cur + 1, cnt + 1, tp, 0); else dfs(cur + 1, cnt, tp, 1); } else if(a[cur] == 9){ if((kg == 1 || kg == 9) && cur != tp - 1 && cur != 0) dfs(cur + 1, cnt + 1, tp, 9); else dfs(cur + 1, cnt, tp, 1); } else dfs(cur + 1, cnt, tp, 1); } int main(){ LL Y; while(~scanf("%lld", &Y)){ int tp = 0; while(Y){ a[tp++] = Y % 10; Y /= 10; } ans = 0; dfs(0, 0, tp, 1); printf("%d\n", tp - ans); } return 0; }
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原文地址：http://www.cnblogs.com/handsomecui/p/5470649.html
