1000. Dice Game





SOJ.Dice Game
20150329 19:33:211000. Dice Game 总提交数量： 100 通过数量： 59 时间限制：1秒 内存限制：256兆 题目描述 Gunnar and Emma play a lot of b题目描述Gunnar and Emma play a lot of board games at home, so they own many dice that are not normal 6 sided dice. For example they own a die that has 10 sides with numbers 47, 48, ..., 56 on it.
There has been a big storm in Stockholm, so Gunnar and Emma have been stuck at home without electricity for a couple of hours. They have finished playing all the games they have, so they came up with a new one. Each player has 2 dice which he or she rolls. The player with a bigger sum wins. If both sums are the same, the game ends in a tie.
Given the description of Gunnar’s and Emma’s dice, which player has higher chances of winning?
All of their dice have the following property: each die contains numbers a, a + 1, ..., b, where a and b are the lowest and highest numbers respectively on the die. Each number appears exactly on one side, so the die has b  a + 1 sides.
输入格式The first line contains four integers a1, b1, a2, b2 that describe Gunnar’s dice. Die number i contains numbers ai, ai + 1, ..., bi on its sides. You may assume that 1 <= ai <= bi <= 100. You can further assume that each die has at least four sides, so ai + 3 <= bi. The second line contains the description of Emma’s dice in the same format.
输出格式Output the name of the player that has higher probability of winning. Output “Tie” if both players have same probability of winning.
样例输入样例一： 1 4 1 4 1 6 1 6 样例二： 1 8 1 8 1 10 2 5 样例三： 2 5 2 7 1 5 2 5
样例输出样例一： Emma 样例二： Tie 样例三： Gunnar
Problem Source: 2015年每周一赛第四场
挺水的一道题，看平均值就好#include using namespace std; int main(){ double a,b,c,d,e,f,g,h,sum1,sum2; cin>>a>>b>>c>>d>>e>>f>>g>>h; sum1=((a+b)/2)+((c+d)/2); sum2=((e+f)/2)+((g+h)/2); if(sum1==sum2)cout<<"Tie"<sum2)cout<<"Gunnar"<

D  Dice Game
20180831 18:06:45A dice is a small cube, with each side having a different number of spots on it, ranging from 1 to 6. Each side in the dice has 4 adjacent sides that can be reached by rotating the dice (...A dice is a small cube, with each side having a different number of spots on it, ranging from 1 to 6.
Each side in the dice has 4 adjacent sides that can be reached by rotating the dice (i.e. the current side) 90 degrees. The following picture can help you to conclude the adjacent sides for each side in the dice.
In this problem, you are given a dice with the side containing 1 spot facing upwards, and a sum n, your task is to find the minimum number of required moves to reach the given sum.
On each move, you can rotate the dice 90 degrees to get one of the adjacent sides to the side that currently facing upwards, and add the value of the new side to your current sum. According to the previous picture, if the side that currently facing upwards contains 1 spot, then in one move you can move to one of sides that contain 2, 3, 4, or 5 spots.
Initially, your current sum is 0. Even though at the beginning the side that containing 1 spot is facing upwards, but its value will not be added to your sum from the beginning, which means that you must make at least one move to start adding values to your current sum.
Input
The first line contains an integer T (1 ≤ T ≤ 200), where T is the number of test cases.
Then T lines follow, each line contains an integer n (1 ≤ n ≤ 104), where n is the required sum you need to reach.
Output
For each test case, print a single line containing the minimum number of required moves to reach the given sum. If there is no answer, print 1.
Example
Input
2 5 10
Output
1 2
Note
In the first test case, you can rotate the dice 90 degrees one time, and make the side that contains 5 spots facing upwards, which make the current sum equal to 5. So, you need one move to reach sum equal to 5.
In the second test case, you can rotate the dice 90 degrees one time, and make the side that contains 4 spots facing upwards, which make the current sum equal to 4. Then rotate the dice another 90 degrees, and make the side that contains 6 spots facing upwards, which make the current sum equal to 10. So, you need two moves to reach sum equal to 10.
#include <iostream> #include<stdio.h> using namespace std; typedef long long ll; int main() {int x,n,t; scanf("%d",&t); while(t) { scanf("%d",&n); if(n==1)printf("1\n"); else if(n==7)printf("3\n"); else { int ans=n/11*2; x=n%11; if(x<=5)ans++; else ans+=2; if(x==0)ans; printf("%d\n",ans); } } return 0; }

Sicily 13907. Dice Game
20150330 22:15:4513907. Dice Game Constraints Time Limit: 1 secs, Memory Limit: 256 MB Description Gunnar and Emma play a lot of board games at home, so they own many dice that are not normal 6 side13907. Dice Game
Constraints
Time Limit: 1 secs, Memory Limit: 256 MB
Description
Gunnar and Emma play a lot of board games at home, so they own many dice that are not normal 6 sided dice. For example they own a die that has 10 sides with numbers 47, 48, ..., 56 on it.
There has been a big storm in Stockholm, so Gunnar and Emma have been stuck at home without electricity for a couple of hours. They have finished playing all the games they have, so they came up with a new one. Each player has 2 dice which he or she rolls. The player with a bigger sum wins. If both sums are the same, the game ends in a tie.
Given the description of Gunnar’s and Emma’s dice, which player has higher chances of winning?
All of their dice have the following property: each die contains numbers a, a + 1, ..., b, where a and b are the lowest and highest numbers respectively on the die. Each number appears exactly on one side, so the die has b  a + 1 sides.
Input
The first line contains four integers a1, b1, a2, b2 that describe Gunnar’s dice. Die number i contains numbers ai, ai + 1, ..., bi on its sides. You may assume that 1 <= ai <= bi <= 100. You can further assume that each die has at least four sides, so ai + 3 <= bi. The second line contains the description of Emma’s dice in the same format.
Output
Output the name of the player that has higher probability of winning. Output “Tie” if both players have same probability of winning.
Sample Input
样例一： 1 4 1 4 1 6 1 6 样例二： 1 8 1 8 1 10 2 5 样例三： 2 5 2 7 1 5 2 5
Sample Output
样例一： Emma 样例二： Tie 样例三： Gunnar
Problem Source
2015年每周一赛第四场
// Problem#: 13907 // Submission#: 3660104 // The source code is licensed under Creative Commons AttributionNonCommercialShareAlike 3.0 Unported License // URI: http://creativecommons.org/licenses/byncsa/3.0/ // All Copyright reserved by Informatic Lab of Sun Yatsen University #include <stdio.h> int main() { int a1, b1, a2, b2, a3, b3, a4, b4; double g = 0, e = 0; scanf("%d%d%d%d%d%d%d%d", &a1, &b1, &a2, &b2, &a3, &b3, &a4, &b4); for (int i = a1; i <= b1; i++) for (int j = a2; j <= b2; j++) g += i + j; for (int i = a3; i <= b3; i++) for (int j = a4; j <= b4; j++) e += i + j; printf(g / ((b1  a1 + 1) * (b2  a2 + 1)) > e / ((b4  a4 + 1) * (b3  a3 + 1)) ? "Gunnar\n" : (g / ((b1  a1 + 1) * (b2  a2 + 1)) < e / ((b4  a4 + 1) * (b3  a3 + 1)) ? "Emma\n" : "Tie\n")); return 0; }

Game of Dice GYM101532E
20171025 19:02:01http://codeforces.com/gym/101532/problem/Emeet in the middle 分成两份 然后后一份最终得到的A需要得到X时的BA*B = X mod (1e9+7) B = X*(inv[A]) mod (1e9+7)#include #include #include ...http://codeforces.com/gym/101532/problem/E
meet in the middle
分成两份 然后后一份最终得到的A需要得到X时的B
A*B = X mod (1e9+7)
B = X*(inv[A]) mod (1e9+7)#include <iostream> #include <algorithm> #include <sstream> #include <string> #include <queue> #include <cstdio> #include <map> #include <set> #include <utility> #include <stack> #include <cstring> #include <cmath> #include <vector> #include <ctime> #include <bitset> using namespace std; #define pb push_back #define sd(n) scanf("%d",&n) #define sdd(n,m) scanf("%d%d",&n,&m) #define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k) #define sld(n) scanf("%lld",&n) #define sldd(n,m) scanf("%lld%lld",&n,&m) #define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k) #define sf(n) scanf("%lf",&n) #define sff(n,m) scanf("%lf%lf",&n,&m) #define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k) #define ss(str) scanf("%s",str) #define ans() printf("%d",ans) #define ansn() printf("%d\n",ans) #define anss() printf("%d ",ans) #define lans() printf("%lld",ans) #define lanss() printf("%lld ",ans) #define lansn() printf("%lld\n",ans) #define fansn() printf("%.10f\n",ans) #define r0(i,n) for(int i=0;i<(n);++i) #define r1(i,e) for(int i=1;i<=e;++i) #define rn(i,e) for(int i=e;i>=1;i) #define rsz(i,v) for(int i=0;i<(int)v.size();++i) #define szz(x) ((int)x.size()) #define mst(abc,bca) memset(abc,bca,sizeof abc) #define lowbit(a) (a&(a)) #define all(a) a.begin(),a.end() #define pii pair<int,int> #define pli pair<ll,int> #define pll pair<ll,ll> #define mp(aa,bb) make_pair(aa,bb) #define lrt rt<<1 #define rrt rt<<11 #define X first #define Y second #define PI (acos(1.0)) #define sqr(a) ((a)*(a)) typedef long long ll; typedef unsigned long long ull; const ll mod = 1000000000+7; const double eps=1e9; const int inf=0x3f3f3f3f; const ll infl = 10000000000000000; const int maxn= 100+10; const int maxm = 6000+10; //Pretests passed int in(int &ret) { char c; int sgn ; if(c=getchar(),c==EOF)return 1; while(c!=''&&(c<'0'c>'9'))c=getchar(); sgn = (c=='')?1:1; ret = (c=='')?0:(c'0'); while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c'0'); ret *=sgn; return 1; } int a[maxn][10]; ll qpow(ll k,ll x) { ll res = 1; ll base = x; while(k) { if(k&1)res=res*base%mod; base = base*base %mod; k>>=1; } return res; } map<ll,int>ma;int n,x; void dfs(int cur,int last,ll now) { if(!last) { ma[now]++; return ; } for(int i=1;i<=6;++i)dfs(cur+1,last  1, now*a[cur][i]%mod); } ll ans = 0; void dfs2(int cur,ll now) { if(cur==n+1) { ll inv = qpow(mod2,now); // ll check = x%now==0?x/now:0; // ll cnt = ma[check]; // ans +=cnt; inv = inv*x%mod ; ll cnt = ma[inv]; ans += cnt; return ; } r1(i,6)dfs2(cur+1,now*a[cur][i]%mod); } int main() { #ifdef LOCAL freopen("input.txt","r",stdin); // freopen("output.txt","w",stdout); #endif // LOCAL int t; sd(t); while(t) { // int n,x; ans = 0; sdd(n,x); r1(i,n)r1(j,6)sd(a[i][j]); ma.clear(); int mid = n/2; dfs(1,mid,1); dfs2(mid+1,1); lansn(); } return 0; }

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