CodeForce #429 DIV2 A B C题解

A:http://codeforces.com/contest/841/problem/A

题意：n个气球分给k个人，是否有这样的解：每个人手里的气球都颜色不重复

思路：个数最多的颜色个球的个数 >k， 就必然有人手里两个球

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;int main(int argc, char *argv[]) {//freopen("A.txt", "r", stdin);int SZ = 26;int n,k;int num[SZ];while(~scanf("%d%d", &n, &k)) {memset(num, 0, sizeof(num));string s;cin >> s;for (int i = 0; i < s.size(); i++) {num[s[i] - 'a'] += 1;}int mx = 0;for (int i = 0; i < SZ; i++) {mx = max(mx, num[i]);//cout << num[i] << " " << mx << endl;}//cout << mx << endl;if (mx > (int)(k))printf("NO\n");elseprintf("YES\n");}return 0;}

B:http://codeforces.com/contest/841/problem/B

题意：一个数组，A可以删除和为奇数的连续子串，B可以删除和为偶数的连续子串，不能操作的输，A是先手，问谁赢

思路：

1、如果和为奇数，A一定赢

2、如果和为偶数，那么有两种情况，一是没有奇数，那么A一定输，如果有奇数，那么奇数必然是偶数个。A作为先手，只能取和为奇数的序列，那么剩下的和必然是奇数，而接下来B只能取和为偶数的序列，B取剩下的就和一定是奇数，那么第二轮A就能取完剩下的所有数字，于是A就赢了。

综合来看，只要有奇数出现，B一定会熟

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cstdlib>using namespace std;int main(int argc, char *argv[]) {int x, n, r;while(~scanf("%d", &n)) {r = 2;for (int i = 0; i < n; i++) {scanf("%d", &x);if(x%2) {r = 1;} }if (r == 1) {printf("First\n");} else {printf("Second\n");}}return 0;}

C:http://codeforces.com/contest/841/problem/C

解很简单，看下example应该都能找到规律。

证明：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int SIZE = 1e5 *2 +1;//int A[SIZE];struct Node {int v;int idx;Node(){}Node(int vv, int i) {v = vv;idx = i;}};Node A[SIZE];Node B[SIZE];bool cmp(const Node a, const Node b) {return a.v > b.v;}bool cmp2(const Node a, const Node b) {return a.v < b.v;}bool cmp3(const Node a, const Node b) {return a.idx < b.idx;}int main() {//freopen("A.txt", "r", stdin);int m;while(~scanf("%d", &m)) {//cout << "m:" << m << endl;for (int i =0; i < m; i++) {scanf("%d", &(A[i].v));A[i].idx = i;}for (int i = 0; i < m; i++) {scanf("%d", &(B[i].v));B[i].idx = i;}sort(A, A+m, cmp);sort(B, B+m, cmp2);for (int i = 0; i < m; i++) {//cout << "i:" << i << " A:" << A[i].v << "," << A[i].idx << " B:"<< B[i].v << "," << B[i].idx << endl;A[i].idx = B[i].idx;} sort(A, A+m, cmp3);printf("%d", A[0].v);for (int i = 1; i < m; i++) {printf(" %d", A[i].v);}putchar('\n');}return 0;}