2018.12.28 22：30
看着CF升高的曲线，摸了摸自己的头发，我以为我变强了，直到这一场Edu搞醒了我。。
从即将进入2018年末开始，开启自闭场集合，以纪念(dian)那些丢掉的头发
留坑睡觉。。明天看题解再补
A.Find Divisible
题意：输出[l,r]中满足x|y的x,y，保证有解
思路：直接输出x, 2x即可
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 1e9+7; const int maxn = 2e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); int main() { int t; scanf("%d", &t); while(t--){ ll x, y; scanf("%lld %lld", &x, &y); printf("%lld %lld ", x,2*x); } return 0; }
题意：删除一个子串，使得剩下的串只有一种字符，问你方案数
思路：分a[1]是否==a[n]两种情况讨论，
等于的时候根据要删除的子串区间端点范围决定方案数
不等于的时候相当于一个端点固定
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 998244353; const int maxn = 2e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); char a[maxn]; int main() { int n; scanf("%d", &n); getchar(); scanf("%s", a+1); ll ans = 0; if(a[1]==a[n]){ char c = a[1]; int l , r; for(int i = 1; i <= n; i++){ if(a[i]!=c){l=i;break;} } for(int i = n; i >= 1; i--){ if(a[i]!=c){r=i;break;} } l = l; r = n-r+1; ans = 1ll*l*r; ans%=mod; } else{ int l,r; char c = a[1]; for(int i = 1; i <= n; i++){ if(a[i]!=c){l=i;break;} } c = a[n]; for(int i = n; i >= 1; i--){ if(a[i]!=c){r=i;break;} } ans = n-r; ans += l; ans%=mod; } printf("%lld", ans); return 0; }
C.Polygon for the Angle
题意：问你一个角度最少存在于正几边形中
思路：看代码。。
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 998244353; const int maxn = 2e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); int main() { int t; scanf("%d", &t); while(t--){ int ag; scanf("%d", &ag); int g = __gcd(ag,180); int n = 180/g; int m = ag/g; while(m>n-2){n*=2;m*=2;} printf("%d ", n); } return 0; }
D.Easy Problem
题意：给你一个字符串，每个字符都有权重，问你删掉多少权重和的字符，使得剩下的字符没有"hard"子序列，并且这个权重和最小
思路：dp[i][j]为前i个字符最多拥有j状态的子序列需要删除的最小权重
其中 j=0代表空，j=1代表"h"，j=2代表"ha"，j=3代表"har"
转移方程在代码里
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 998244353; const int maxn = 2e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); char s[maxn]; ll dp[maxn][6]; //dp[i][j]表示前i个字符最多有prej作为子序列要删多少 int a[maxn]; int n; int main() { scanf("%d", &n); getchar(); scanf("%s", s+1); for(int i = 1; i <= n; i++){ scanf("%d", &a[i]); } mem(dp,0); //dp[0][0] = 0; for(int i = 1; i <= n; i++){ for(int j = 0; j < 4; j++){ dp[i][j] = dp[i-1][j]; } if(s[i]=='h')dp[i][0]+=a[i]; if(s[i]=='a')dp[i][1]=min(dp[i-1][0],dp[i][1]+a[i]); if(s[i]=='r')dp[i][2]=min(dp[i-1][1],dp[i][2]+a[i]); if(s[i]=='d')dp[i][3]=min(dp[i-1][2],dp[i][3]+a[i]); } ll ans = 0x7f7f7f7f7f7f7f7f; for(int i = 0; i <= 3 ;i++){ ans = min(ans, dp[n][i]); }printf("%lld", ans); return 0; }