Coder-Strike 2014 - Finals Div 2
2014-04-23 17:41
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B题:
输入一个只含‘A’-‘Z’的字符串,要求每个字符是对称的且是回文串。
C题:
题意:输入n和p,接下来有n行,第i行两个a,b,代表第i个人认为a和b是嫌疑犯。领导要从中选出两个嫌疑犯,问有多少种选法使至少有p个人同意他的决定。某个人同意定义为这两个人中至少有一个人是他认为的嫌疑犯。
思路:拿一个数组存每个嫌疑犯被怀疑的次数,升序排序,对第i 个嫌疑犯,二分查找[i+1,n]到第j个人满足i和j被怀疑的次数之和大于等于p,那么j之后的与i之和肯定大于等于p。求得ans。但结果偏大,因为可能有同一个人怀疑i和j。那么要更新ans。
学习:
lower_bound():
函数lower_bound()在first和last中的前闭后开区间进行二分查找,返回大于或等于val的第一个元素位置。如果所有元素都小于val,则返回last的位置
map和pair的应用。
输入一个只含‘A’-‘Z’的字符串,要求每个字符是对称的且是回文串。
#include <iostream> #include <string> #include <cmath> #include <math.h> #include <queue> #include <algorithm> #include <stack> #include <vector> #include <string.h> #include <stdio.h> using namespace std; int hash[26]; void init() { memset(hash,0,sizeof(hash)); for(int i = 0; i < 26; i++) { if(i+'A' == 'A' || i+'A' == 'H' || i+'A' == 'I' || i+'A' == 'M' || i+'A' == 'O' || i+'A' == 'T' || i+'A' == 'U' || i+'A' == 'V' || i+'A' == 'W' || i+'A' == 'X' || i+'A' == 'Y') hash[i] = 1; } } int main() { char s[100010]; int len; init(); while(~scanf("%s",s)) { len = strlen(s); int flag = 1; for(int i = 0; i < len; i++) { if(s[i] != s[len-1-i] || hash[s[i]-'A'] == 0) { flag = 0; break; } } if(flag == 0) printf("NO\n"); else printf("YES\n"); } return 0; }
C题:
题意:输入n和p,接下来有n行,第i行两个a,b,代表第i个人认为a和b是嫌疑犯。领导要从中选出两个嫌疑犯,问有多少种选法使至少有p个人同意他的决定。某个人同意定义为这两个人中至少有一个人是他认为的嫌疑犯。
思路:拿一个数组存每个嫌疑犯被怀疑的次数,升序排序,对第i 个嫌疑犯,二分查找[i+1,n]到第j个人满足i和j被怀疑的次数之和大于等于p,那么j之后的与i之和肯定大于等于p。求得ans。但结果偏大,因为可能有同一个人怀疑i和j。那么要更新ans。
学习:
lower_bound():
函数lower_bound()在first和last中的前闭后开区间进行二分查找,返回大于或等于val的第一个元素位置。如果所有元素都小于val,则返回last的位置
map和pair的应用。
#include <stdio.h> #include <algorithm> #include <set> #include <map> #include <vector> #include <string.h> #define LL long long #define _LL __int64 using namespace std; const int INF = 0x3f3f3f3f; map < pair<int, int>, int > mp; int n,p; int c1[300010],c2[300010]; _LL ans; int main() { int i,j,a,b; scanf("%d %d",&n,&p); memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); for(i = 1; i <= n; i++) { scanf("%d %d",&a,&b); if(a > b) swap(a,b); c1[a]++; c1[b]++; c2[a]++; c2[b]++; mp[ make_pair(a,b) ] += 1; } sort(c1+1,c1+1+n); map< pair<int,int>,int>::iterator it; ans = 0; for(i = 1; i <= n; i++) { int add = lower_bound(c1+1+i,c1+1+n,p-c1[i]) - c1; //lower_bound函数进行二分查找第一个大于等于p-c[i]的位置 ans += n+1-add; } for(it = mp.begin(); it != mp.end(); it++) { a = it->first.first; b = it->first.second; if(c2[a]+ c2[b] >= p && c2[a] + c2[b] - mp[make_pair(a,b)] < p) //不满足的要减去 ans--; } printf("%I64d\n",ans); return 0; }
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