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HDU - 6034 Balala Power!

题意:有n个字符串,为出现过的每种字母赋一个0~25的值,要求每种字母的值各不相同。要求,若一个字母是一个长度大于1的字符串的首字母,则该字母不能为0。为所有字母赋值后,所得到的每个字符串都是二十六进制数,问这些二十六进制数转化为十进制后的数值之和。

分析:

1、统计出现过的每种字母在每一位上的出现次数。若某一位的出现次数>=26,则进位。

2、按照处理过的各字母在每一位上的出现次数排序,高位出现次数多的优先级高,并依次为每个字母赋值。

3、若优先级最低的字母j不能赋值为0,则从优先级由低到高,找到第一个可以赋值为0的字母i将其赋值为0,并将i之后的字母的赋值数+1。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const LL MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 100000 + 10;const int MAXT = 1000000 + 10;using namespace std;char s[MAXN];int cnt[30][MAXN];int maxlen;set<int> st;struct Node{    int c;    bool leadingzero;    int value;    bool operator < (const Node& rhs)const{        for(int i = maxlen - 1; i >= 0; --i){            if(cnt[c][i] != cnt[rhs.c][i]) return cnt[c][i] > cnt[rhs.c][i];        }        return c < rhs.c;    }}num[30];int main(){    int n;    int kase = 0;    while(scanf("%d", &n) == 1){        memset(cnt, 0, sizeof cnt);        st.clear();        maxlen = 0;        for(int i = 0; i < 26; ++i) num[i].c = i, num[i].leadingzero = false;        for(int i = 0; i < n; ++i){            scanf("%s", s);            int len = strlen(s);            if(len > 1) num[s[0] - ‘a‘].leadingzero = true;            maxlen = max(maxlen, len);            for(int j = 0; j < len; ++j){                ++cnt[s[j] - ‘a‘][len - 1 - j];                st.insert(s[j] - ‘a‘);            }        }        for(int i = 0; i < 26; ++i){            for(int j = 0; j < maxlen - 1; ++j){                while(cnt[i][j] >= 26){                    cnt[i][j] -= 26;                    ++cnt[i][j + 1];                }            }        }        sort(num, num + 26);        for(int i = 0; i < st.size(); ++i){            num[i].value = http://www.mamicode.com/25 - i;"Case #%d: %lld\n", ++kase, ans);    }    return 0;}

  

HDU - 6034 Balala Power!