2017 Multi-University Training Contest
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6034
题目大意: 给你n个字符串, a~z的每个字母可以映射为0~25的一个数字, 每个字符串会形成一个26进制的数, 让所有的字符串加起来的和最大, 并且不能有前导零。
解题思路: 对于每一个字母, 他都有一个位权值, 只需要把所有的字母的位权和从大到小排序并且一一映射成数字, 则是最优的, 当然还要考虑最后0是否会出现前导零, 如果有, 就需要往前找最近的一个不出现前导零的。
#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#include<queue>using namespace std;typedef long long LL;const int MaxN = 2e5, pt = 1e9 + 7;int num[26][MaxN + 5];LL power[MaxN + 5], sum[MaxN + 5];bool ban[26];int a[26];char str[MaxN + 5];int n, L, Case;bool cmp(int a, int b){for(int i = L - 1; i >= 0; i--){if(num[a][i] != num[b][i]) return num[a][i] < num[b][i];}return 0;}void solve(){memset(num, 0, sizeof(num));memset(sum, 0, sizeof(sum));memset(ban, 0, sizeof(ban));L = 0;for(int i = 1; i <= n; i++){scanf("%s", 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];while(sum[str[j] - 'a'] > pt) sum[str[j] - 'a'] -= pt;}L = max(L, len);}for(int i = 0; i < 26; i++) //对权值进位 { for(int j = 0; j < L; j++){ num[i][j + 1] += num[i][j] / 26; num[i][j] %= 26; } while (num[i][L]){//将最高位向后拓展 num[i][L + 1] += num[i][L] / 26; num[i][L++] %= 26; } 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;}}int res = 0, x = 25;for(int i = 25; i >= 0; i--){if(a[i] != zero){res = (res + x * sum[a[i]] % pt) % pt;x--;}}printf("Case #%d: %d\n", ++Case, res);}int main(){power[0] = 1LL;for(int i = 1; i <= MaxN; i++) power[i] = 26LL * power[i - 1] % pt;while(~scanf("%d", &n)){solve();}return 0;}
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- 2017 Multi-University Training Contest
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- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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- 2017 Multi-University Training Contest
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