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Three players, A, B, and C play a dice game, each player rolls two dice.

（The points of 2*3 = 6 dice should be set and stored in advance)

PART I:

Assume that each player can only see the points of his/her own two dice, simulate their game process.

Rules of PART I:

Starting from player A, the following steps are performed in turn.

First, the two dice thrown by the player A are output (indicating that A sees his/her dice), and then two numbers a and b are required to be input to represent A's guess for all dice, which means the number of b point dices in total is at least a.

Then it's the next player B's turn. After seeing his own dice, B can choose to enter 0 0 (two zeros) for "I don't believe" or to enter a new pair of numbers c and d, which means he/she guesses "there are at least c d points in all dice", where c,d must satisfy c≥a, and when c=a, d>b.

The game continues until someone input 0 0 (two zeros).

Optional rules：1 point dice can be regarded as any number of points.

For example:

O: A's turn: A has 2 and 1
I: 2 4
O: B's turn: B has 3 and 2
I: 2 3
O: Invalid input.
O: B's turn: B has 3 and 2
I: 3 2
O: C's turn: C has 2 and 5
I: 3 5
O: A's turn: A has 2 and 1
I: 4 2
O: B's turn: B has 3 and 2
I: 0 0

If someone (let's say player B above) enters 0 0 (two zeros), then show all 2 * 3 = 6 dice, determine the current player's "unbelieving" is right or wrong to judge the loser of the game: Assume that the previous player's input is x y. If the number of dice in y points is less than x (in a total of 6 dice), then the previous player (A) loses, otherwise, the current player (B) loses.

For example:

O: A's turn: A has 2 and 1
I: 2 4
O: B's turn: B has 3 and 2
I: 2 3
O: Invalid input.
O: B's turn: B has 3 and 2
I: 3 2
O: C's turn: C has 2 and 5
I: 3 5
O: A's turn: A has 2 and 1
I: 4 2
O: B's turn: B has 3 and 2
I: 0 0
O: All dices: 2 1 3 2 2 5
Loser is A. (Without optional rules)

Loser is B. (With optional rules)

PART II:

Play this game from the perspective of God, i.e. you can see all of the 6 dice.

Rules of PART II:

Observe all 6 dice and output the optimal combination of a and b, which means it  will certainly  be wrong for any other player to guess again.

For example:

O: All dices: 2 1 3 2 2 5
Best answer: 3 2 (Without optional rules)

Best answer: 4 2 (With optional rules)

注：

O: 用例输出，I: 用例输入，不必写在程序输出里

要求体现代码的良好风格、友好界面、算法和代码的高执行效率等。

其他可选部分：例如，可选择2-5之间的数字作为游戏人数及骰子个数等。

开放时间:    2018年04月02日星期一 12:00
截止时间:    2018年04月21日星期六 00:00
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#include<stdio.h> #include<stdlib.h> #include<time.h> int checknum(int a,int b,int c,int d) { if(a>6||b>6||c>6||d>6||a<0||b<0||c<0||d<0) return 1; if(c==0&&d==0) return 0; if(c>a||(c==a&&d>b)) return 0; else return 1; } void catchlier(int a,int b,int *p,int o) { int i,m=0,n=0; for(i=0;i<6;++i,*(p++)) { if(*p==1) m++; if(*p==b) n++; } if(n>=a) printf("Loser is %c. (Without optional rules)\n",o); else printf("Loser is %c. (Without optional rules)\n",o>65?o-1:67); if(m+n>=a) printf("Loser is %c. (With optional rules)\n",o); else printf("Loser is %c. (With optional rules)\n",o>65?o-1:67); } int main(void) { int points[6],i,n=65,j=0; int a,b,c,d; srand((unsigned)time(NULL)); for(i=0;i<6;++i) points[i]=rand()%6+1; printf("A's turn: A has %d and %d\n",points[0],points[1]); scanf("%d%d",&a,&b); while(1) { j++; n++; if(n>67) n=65; if(j>2) j=1; printf("%c's turn: %c has %d and %d\n",n,n,points[j*2],points[j*2+1]); scanf("%d%d",&c,&d); while(checknum(a,b,c,d)) { printf("Invalid input.\n"); printf("%c's turn: %c has %d and %d\n",n,n,points[j*2],points[j*2+1]); scanf("%d%d",&c,&d); } if(c==0&&d==0) break; a=c; b=d; } printf("All dices:"); for(i=0;i<6;++i) printf("%d ",points[i]); putchar('\n'); catchlier(a,b,points,n); return 0; }