hdu 4291 && hdu 4296
题目:http://acm.hdu.edu.cn/showproblem.php?pid=4291
题意和解释网上一搜一大片,以前也做过一个用矩阵乘法求这种题的(这个还要求出嵌套的循环节),可是看到题的时候还是什么也没有想到,对以前的知识掌握的不好,也不会灵活用。。
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1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <math.h> 7 #define _clr(a,val) (memset(a,val,sizeof(a))) 8 9 using namespace std; 10 11 typedef long long ll; 12 const ll mod1 = 183120; 13 const ll mod2 = 222222224; 14 const ll mod3 = 1000000007; 15 ll mod; 16 /*void tcal(ll mod) 找循环节 17 { 18 ll i; 19 ll tem = 0, kem = 1, tt = 0; 20 i = 1; 21 while(1) 22 { 23 tt = (3 * kem % mod + tem % mod) % mod; 24 tem = kem; 25 kem = tt; 26 if(tem == 0 && kem == 1) 27 { 28 cout<<i<<endl; 29 break; 30 } 31 i++; 32 } 33 }*/ 34 struct node 35 { 36 ll a[2][2]; 37 void init() 38 { 39 for(int i = 0; i < 2; i++) 40 { 41 for(int j = 0; j < 2; j++) 42 a[i][j] = (i == j); 43 } 44 } 45 }; 46 node cal(node t,node b) 47 { 48 node c; 49 _clr(c.a,0); 50 int i,j,k; 51 for(k = 0; k < 2; k++) 52 { 53 for(i = 0; i < 2; i++) 54 { 55 for(j = 0; j < 2; j++) 56 { 57 c.a[i][j] += ((t.a[i][k] * b.a[k][j]) % mod); 58 } 59 } 60 } 61 for(i = 0; i < 2; i++) 62 { 63 for(j = 0; j < 2; j++) 64 c.a[i][j] %= mod; 65 } 66 return c; 67 } 68 ll ca(node t,ll n,ll tmod) 69 { 70 node c; 71 c.init(); 72 mod = tmod; 73 while(n) 74 { 75 if(n & 1) c = cal(c,t); 76 t = cal(t,t); 77 n /= 2; 78 } 79 return c.a[0][1]; 80 } 81 int main() 82 { 83 ll n; 84 //tcal(mod3); // 找出 10 ^ 9 + 7 的循环节 85 //tcal(mod2); // 找出 (10^9+7)的循环节的循环节 86 //freopen("data.txt","r",stdin); 87 while(cin>>n) 88 { 89 if(n < 0) break; 90 node t; 91 t.a[0][0] = 3, t.a[0][1] = 1; 92 t.a[1][0] = 1, t.a[1][1] = 0; 93 ll ans = ca(t,ca(t,ca(t,n,mod1),mod2),mod3); 94 printf("%lld\n",ans); 95 } 96 return 0; 97 }
题目:http://acm.hdu.edu.cn/showproblem.php?pid=4296
这个题悲剧就是我理解错了,然后给他们也说错了,悲剧的到最后才理解题目的真正意思
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1 typedef long long ll; 2 const int N = 100010; 3 struct node 4 { 5 int w; 6 int h; 7 }b[N]; 8 int cmp(node a,node b) 9 { 10 return (a.w + a.h) > (b.w + b.h); 11 } 12 int main() 13 { 14 int i,j; 15 ll tsum; 16 int n; 17 //freopen("data.txt","r",stdin); 18 while(scanf("%d",&n) != EOF) 19 { 20 tsum = 0; 21 for(i = 0; i < n; i++) 22 { 23 scanf("%d%d",&b[i].w,&b[i].h); 24 tsum += b[i].w; 25 } 26 sort(b, b + n,cmp); 27 ll phd = 0; 28 ll maxx = 0; 29 for(j = 0; j < n - 1; j++) 30 { 31 tsum -= b[j].w; 32 phd = (tsum - b[j].h); 33 if(maxx < phd) maxx = phd; 34 } 35 cout<<maxx<<endl; 36 } 37 return 0; 38 }