Codeforces Round #422 (Div. 2)

A

注意A、B的条件即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 2e5 + 10;
int a[qq], b[qq];

int main(){
	int A, B;	scanf("%d%d", &A, &B);
	LL ans = 1;
	for(int i = 1; i <= min(A, B); ++i)
		ans *= (LL)i;
	printf("%lld\n", ans);
	return 0;
}

B

题意：给出两个字符串s、t，问把s中某些字符改变成？使得字符串s是字符串t的某个字串，问改变字符最少的是多少

思路：枚举s串是t串中某个字串，记录最小改变次数即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 2e5 + 10;
int n, m;
char s[1005], t[1005];
map<string, int> mp;
int main(){
	scanf("%d%d", &n, &m);
	scanf("%s%s", s + 1, t + 1);
	int lens = strlen(s + 1), lent = strlen(t + 1);
	int l = 0, r = 0, minx = 1e9;
	for(int i = lens; i <= lent; ++i){
		int cnt = 0;
		for(int j = i - lens + 1, k = 1; j <= i; ++j, ++k)
			if(s[k] != t[j]){
				cnt++;
			}
		if(cnt < minx){
			minx = cnt;
			l = i - lens + 1, r = i;
		}
	}
	printf("%d\n", minx);
	for(int i = l, k = 1; i <= r; ++i, ++k){
		if(s[k] != t[i]){
			printf("%d ", k);
		}
	}
	puts("");
	return 0;
}

C

题意：n个区间，l、r、c，问选择两个不相交的区间使得c的和最小，并且两个区间长度的和等于x

思路：这类思维题其实碰到过几次，每次就是想不到点上去，可以这样想，每次如果我确定的某个区间的l值，那么我能选择的就是比l小的区间r值，这样一定能保证选到答案

确定的l，然后更新r。

#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 2e5 + 10, INF = 2e9 + 10;
vector<pill> a[qq], b[qq];
int n, x, cost[qq];

int main(){
	scanf("%d%d", &n, &x);
	for(int i = 0; i < n; ++i){
		int l, r, c;	scanf("%d%d%d", &l, &r, &c);
		a[l].pb({r - l + 1, c});
		b[r].pb({r - l + 1, c});
	}
	fill(&cost[0], &cost[0] + qq, INF);
	int ans = INF;
	for(int i = 0; i < qq; ++i){
		for(auto p : a[i]){
			if(p.first >= x)	continue;
			if(cost[x - p.first] < INF)
				ans = min(ans, p.second + cost[x - p.first]);
		}
		for(auto p : b[i])	cost[p.first] = min(cost[p.first], p.second);
	}
	if(ans == INF)	puts("-1");
	else	printf("%d\n", ans);
	return 0;
}