2017 Multi-University Training Contest
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题目链接:Balala Power!
题目大意:给出的字符串,每个字符建立一种与0-25的对应关系。然后每个字符串看成是一个26进制的数。问能获得的数的总和的最大值。(最后对1e9+7取模)。
题解:每个字符对答案的贡献都可以看作一个 26 进制的数字,问题相当于要给这些贡献加一个 0 到 25 的权重使得答案最大。
做法:按字典序排序过后,贪心。
坑点:除了单独‘0’外,每个数字串不能有前导0;题目保证输入数据至少有一个字符不会出现在任何字符串的开头。(因此,当不得不把某个字符置为0的情况需处理,请看代码)
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<vector>#include<map>#include<queue>#include<stack>typedef long long ll;using namespace std;const int maxn = 1e5 + 10;const ll MOD = 1e9 + 7;int num[26][maxn];int ban[26];char str[maxn];ll power[maxn];ll sum[maxn];int a[26];int n;int maxL;void init(){ memset(ban,0,sizeof(ban)); memset(num,0,sizeof(num)); memset(sum,0,sizeof(sum)); maxL = 0;}bool cmp(const int a,const int b){ for(int i = maxL - 1; i >= 0; i--){ if(num[a][i] != num[b][i]){ return num[a][i] < num[b][i]; } } return 0;}void solve(){ init(); for(int i = 0; i < n; i++){ cin >> str; int len = strlen(str); if(len > 1) ban[str[0] - 'a'] = 1; reverse(str,str+len); for(int j = 0 ; j < len; j++){ num[str[j] - 'a'][j]++; sum[str[j] - 'a'] += power[j]; if(sum[str[j] - 'a'] > MOD){ sum[str[j] - 'a'] %= MOD; } } maxL = max(maxL,len); } for(int i = 0; i < 26; i++){ for(int j = 0; j < maxL; j++){ num[i][j+1] += num[i][j] / 26; num[i][j] %= 26; } while(num[i][maxL]){ num[i][maxL+1] += num[i][maxL] / 26; num[i][maxL] %= 26; maxL++; } a[i] = i; } sort(a,a+26,cmp); int zero = -1; for(int i = 0; i < 26 ; i++){ if(!ban[a[i]]){ zero = a[i]; break; } }//当不得不把某个字符置为0的时候,找到一个最优解:即能使某字符为0的最优策略,当然是选择字典序尽量小的了。 ll ans = 0; int weight = 25; for(int i = 25; i >= 0 ; i--){ if(a[i] != zero){ ans += (ll)(weight--) * sum[a[i]] % MOD; ans %= MOD; } } static int cas = 1; printf("Case #%d: %lld\n",cas++,ans);}int main(){ ios::sync_with_stdio(false); power[0] = 1; for(int i = 1; i < maxn; i++){ power[i] = (power[i-1] * 26) % MOD; } while(cin >> n){ solve(); } return 0;}
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