本文主要是介绍Codeforces 402A 402B 402C 402D，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
402A 直接暴力
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;int main()
{int k, a, b, v;scanf("%d%d%d%d", &k, &a, &b, &v);int ans = 0;while(a > 0){ans ++;if(b >= k-1){a -= k*v;b = b-k+1;}else{a -= (b+1)*v;b = 0;}}printf("%d\n", ans);return 0;
}402B
修改尽量少的数字使数列成为公差为k的等差数列 （但是不能有负数）
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;int num[1010], ans[1010], tem[1010];
int main()
{int n, k;memset(ans, 0, sizeof(ans));scanf("%d%d", &n, &k);for(int i = 0; i < n; ++ i){scanf("%d", &num[i]);tem[i] = num[i] - i*k;}sort(tem, tem+n);int j = 0;while(tem[j] <= 0) ++ j;int now = tem[j], _max = 0, cnt = 0, cc = tem[j];for(int i = j; i < n; ++ i){if(now == tem[i]) ++ cnt;else{if(cnt > _max){_max = cnt;cc = now;}now = tem[i];cnt = 1;}}if(cnt > _max){_max = cnt;cc = now;}printf("%d\n", n-_max);for(int i = 0; i < n; ++ i){if(cc+i*k > num[i]) printf("+ %d %d\n", i+1, cc+i*k-num[i]);if(cc+i*k < num[i]) printf("- %d %d\n", i+1, num[i]-cc-i*k);}return 0;
}402C 猜了一下 直接枚举一个最有可能的
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;int main()
{int t, n, p;scanf("%d", &t);while(t --){scanf("%d%d", &n, &p);int now = 2*n+p, st = 1, k = 1;while(now--){printf("%d %d\n", st, (st+k)%n == 0 ? n : (st+k)%n);++ st;if(st == n+1){st = 1;++ k;}}}return 0;
}402D
数学加贪心
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;int a[5010], cou[5010], pp[100000], m;
bool vis[100000];
struct node
{int b[5050];bool find(int x){for(int i = 0; i < m; ++ i)if(b[i] == x) return true;return false;}
};
node prime;int fi_pri()
{int j = 0;memset(vis, 1, sizeof(vis));for(ll i = 2; i < 100000; ++ i)if(vis[i]){pp[j++] = i;for(ll j = i*i; j < 100000; j += i)vis[j] = false;}return j;
}int gcd(int r, int l)
{return l == 0 ? r : gcd(l, r%l);
}bool can(int i, int j)
{int ans = 0;for(int k = 0; k < j; ++ k)if((i % pp[k] == 0) && (prime.find(pp[k])))while(i % pp[k] == 0){-- ans;i /= pp[k];}elsewhile(i % pp[k] == 0){++ ans;i /= pp[k];}if(prime.find(i)) --ans;if(ans < 0) return true;return false;
}int f(int s, int j)
{if(s == 1) return 0;int i;for(i = 0; i < j; ++ i)if(s%pp[i] == 0)break;if(i != j && prime.find(pp[i]))return f(s/pp[i], j) -1;if(i != j)return f(s/pp[i], j) +1;if(prime.find(s))return -1;return 1;
}int main()
{int asd = fi_pri();int n, g;scanf("%d%d", &n, &m);scanf("%d", &a[0]);g = cou[0] = a[0];for(int i = 1; i < n; ++ i){scanf("%d", &a[i]);g = cou[i] = gcd(g, a[i]);}for(int i = 0; i < m; ++ i)scanf("%d", &prime.b[i]);for(int i = n-1; i >= 0; -- i)if(can(cou[i], asd))for(int j = 0; j <= i; ++ j){a[j] /= cou[i];cou[j] /= cou[i];}int ans = 0;for(int i = 0; i < n; ++ i) ans += f(a[i], asd);printf("%d\n", ans);return 0;
}
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