[CF805D] Minimum number of steps(找规律,贪心)
2017-05-15 16:08
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题目链接:http://codeforces.com/contest/805/problem/D
题意:给一个字符串只有a、b,给一个操作ab->bba,现在希望让字符串不包含ab。问多少次操作?
手动模拟下发现倒着来会得到最优解,每次ab->bba 操作会让b多一个,这会影响前面的a。考察最靠右的a,它移到最右侧需要b的个数步,并且在这个过程中,会将一个b变成两个,这下结论就明显了。
题意:给一个字符串只有a、b,给一个操作ab->bba,现在希望让字符串不包含ab。问多少次操作?
手动模拟下发现倒着来会得到最优解,每次ab->bba 操作会让b多一个,这会影响前面的a。考察最靠右的a,它移到最右侧需要b的个数步,并且在这个过程中,会将一个b变成两个,这下结论就明显了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long LL; 5 const LL mod = (LL)1e9+7; 6 const int maxn = 1001000; 7 char s[maxn]; 8 int n; 9 10 int main() { 11 // freopen("in", "r", stdin); 12 while(~scanf("%s", s)) { 13 n = strlen(s); 14 LL ret = 0; 15 LL cnt = 0; 16 for(int i = n - 1; i >= 0; i--) { 17 if(s[i] == 'a') { 18 ret = (ret + cnt) % mod; 19 cnt = (cnt * 2) % mod; 20 } 21 else cnt = (cnt + 1) % mod; 22 } 23 cout << ret << endl; 24 } 25 return 0; 26 }
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