2017.10.26 water question Major combat part of question solution, 2017.10.26 question Solution

2017.10.26 water question Major combat part of question solution, 2017.10.26 question Solution

 

I feel that this topic is beyond the outline of QWQ...

QWQ ......

A P2907 [usaco 08open] Roads Around The Farm

Why do I feel this question is not as difficult as getting started ..

This is my idea.

For each n, k

Solving a equations makes

X-y = kx −y = k

X + y = nx + y = n

Then perform similar operations on the obtained x and y respectively.

It looks simple,

But I am stuck with QWQ by precision ....

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <stdlib. h> 6 # include <ctime> 7 using namespace std; 8 const int MAXN = 0x7fffff; 9 const int INF = 50; 10 inline int read () 11 {12 char c = getchar (); int f = 1, x = 0; 13 while (c <'0' | c> '9 ') {if (c = '-') f =-1; c = getchar ();} 14 while (c> = '0' & c <= '9 ') x = x * 10 + c-48, c = getchar (); return x * f; 15} 16 int ans = 0; 17 void dfs (int n, int k) 18 {1 9 double p = (double) (n + k)/2; 20 double q = (double) n-(n + k)/2; 21 int o = (n + k)/2; 22 double l = p-(double) o; 23 if (l! = 0 | p> = n & q <= 0 | q> = n & p <= 0) 24 {25 ans ++; 26 return; 27} 28 dfs (p, k); dfs (q, k); 29} 30 int main () 31 {32 int n = read (), k = read (); 33 dfs (n, k); 34 printf ("% d", ans); 35 return 0; 36}View CodeB P1851 good friend

Affinity

Direct brute-force computing

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <stdlib. h> 6 # include <ctime> 7 # define LL long 8 using namespace std; 9 const ll maxn = 0x7fffff; 10 const ll inf = 50; 11 inline LL read () 12 {13 char c = getchar (); LL f = 1, x = 0; 14 while (c <'0' | c> '9 ') {if (c = '-') f =-1; c = getchar ();} 15 while (c> = '0' & c <= '9 ') x = x * 10 + c-48, c = getchar (); return x * f; 16} 17 int main () 18 {19 LL N = read (); 20 while (n ++) 21 {22 LL p = n; 23 LL tot = 0, tot2 = 0; 24 for (LL I = 1; I <p; I ++) if (p % I = 0) tot + = I; 25 for (LL I = 1; I <tot; I ++) if (tot % I = 0) tot2 + = I; 26 if (p = tot2 & p! = Tot) 27 {printf ("% lld", p, tot); exit (0) ;}28 29} 30 return 0; 31}View CodeC P1926 bookboy

01 backpack, calculate the maximum value and release time

Greedy calculation for the most frequently asked questions

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <stdlib. h> 6 # include <ctime> 7 # include <algorithm> 8 using namespace std; 9 const int MAXN = 151; 10 const int INF = 50; 11 inline int read () 12 {13 char c = getchar (); int f = 1, x = 0; 14 while (c <'0' | c> '9 ') {if (c = '-') f =-1; c = getchar ();} 15 while (c> = '0' & c <= '9 ') x = x * 10 + c-48, c = getchar (); return x * f; 16} 17 int n, m, r, k; 18 int dp [MAXN] [MAXN]; 19 // 0 time 20 // 1 score 21 int ti [MAXN]; 22 struct node23 {24 int time; 25 int val; 26} hw [MAXN]; 27 int main () 28 {29 n = read (); m = read (); k = read (); r = read (); 30 for (int I = 1; I <= n; I ++) ti [I] = read (); 31 for (int I = 1; I <= m; I ++) hw [I]. time = read (); 32 for (int I = 1; I <= m; I ++) hw [I]. val = read (); 33 for (int I = 1; I <= m; I ++) 34 for (int j = 0; j <= r; j ++) 35 if (j  

D P1496 burning Chibi

Sort by left endpoint

Maintain the leftmost and rightmost values

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <stdlib. h> 6 # include <ctime> 7 # include <algorithm> 8 # define LL long 9 using namespace std; 10 const ll maxn = 20001; 11 const ll inf = 50; 12 inline LL read () 13 {14 char c = getchar (); LL f = 1, x = 0; 15 while (c <'0' | c> '9') {if (c = '-') f =-1; c = getchar ();} 16 while (c> = '0' & c <= '9') x = x * 10 + c-48, c = getchar (); return x * f; 17} 18 struct node19 {20 LL bg, ed; 21} a [MAXN]; 22 LL n; 23 LL comp (const node & a, const node & B) 24 {25 return. bg <B. bg; 26} 27 LL ans; 28 LL nowleft, nowright; 29 int main () 30 {31 n = read (); 32 for (LL I = 1; I <= n; I ++) 33 {34 a [I]. bg = read (), a [I]. ed = read (); 35 if (a [I]. bg> a [I]. ed) swap (a [I]. bg, a [I]. ed); 36} 37 38 sort (a + 1, a + n + 1, comp); 39 nowleft = a [1]. bg, nowright = a [1]. ed; 40 a [n + 1]. bg = 1e10 + 10; 41 a [n + 1]. ed = 1e10 + 20; 42 for (LL I = 2; I <= n + 1; I ++) 43 {44 if (a [I]. bg <= nowright) nowright = max (nowright, a [I]. ed); 45 else46 {47 ans + = abs (nowright-nowleft); 48 nowleft = a [I]. bg; nowright = a [I]. ed; 49} 50} 51 printf ("% lld", ans); 52 return 0; 53}View Code

 

E P1302 visible rectangle will not do hahaha

 

 

 

 

 

 

# Include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <stdlib. h >#include <ctime> # define LL long usingnamespacestd; const ll maxn = 0x7fffff; const ll inf = 50; inline LL read () {char c = getchar (); LL f = 1, x = 0; while (c <'0' | c> '9') {if (c = '-') f =-1; c = getchar ();} while (c> = '0' & c <= '9') x = x * 10 + c-48, c = getchar (); return x * f;} int main () {LL n = read (); while (n ++) {LL p = n; LL tot = 0, tot2 = 0; for (L L I = 1; I <p; I ++) if (p % I = 0) tot + = I; for (LL I = 1; I <tot; I ++) if (tot % I = 0) tot2 + = I; if (p = tot2 & p! = Tot) {printf ("% lld", p, tot); exit (0) ;}} return0 ;}