UVA11489Integer Game博弈
2015-07-03 16:23
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题目大意:
S和T在玩游戏,S先。给出一数字串,两人轮流取出一个数字,要求每次取完之后剩下的数为3的倍数,或者没有数字留下。如果两个人足够聪明,求胜利的一方。
对于本题而言,S先手,他若想赢,只需要处理好两种情况1.总和为3的倍数2.总和不是3的倍数.
对于1.S会选择第一次拿一个3,6,9,轮到T也想拿3,6,9这时需要数总共多少个3,6,9总共奇数个则S赢
对于2.S必须拿一个非3,6,9而且使剩下的凑成3的倍数。这时T面临的情况就和S在1.开始时相同,而S若想赢,3,6,9的总和就得是偶数
11489 Integer GameTwo players, S and T, are playing a game where they make alternate moves. S plays first.In this game, they start with an integer N. In each move, a player removes one digit from theinteger and passes the resulting number to the other player.
The game continues in this fashion untila player finds he/she has no digit to remove when that player is declared as the loser.With this restriction, its obvious that if the number of digits in N is odd then S wins otherwise Twins. To make the game more interesting,
we apply one additional constraint. A player can remove aparticular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left.Suppose N = 1234. S has 4 possible moves. That is, he can remove 1, 2, 3, or 4. Of these,
two ofthem are valid moves.• Removal of 4 results in 123 and the sum of digits = 1 + 2 + 3 = 6; 6 is a multiple of 3.• Removal of 1 results in 234 and the sum of digits = 2 + 3 + 4 = 9; 9 is a multiple of 3.The other two moves are invalid.If both players play
perfectly, who wins?InputThe first line of input is an integer T (T < 60) that determines the number of test cases. Each case isa line that contains a positive integer N. N has at most 1000 digits and does not contain any zeros.OutputFor each case, output
the case number starting from 1. If S wins then output ‘S’ otherwise output ‘T’.Sample Input3433771Sample OutputCase 1: SCase 2: TCase 3: T
S和T在玩游戏,S先。给出一数字串,两人轮流取出一个数字,要求每次取完之后剩下的数为3的倍数,或者没有数字留下。如果两个人足够聪明,求胜利的一方。
对于本题而言,S先手,他若想赢,只需要处理好两种情况1.总和为3的倍数2.总和不是3的倍数.
对于1.S会选择第一次拿一个3,6,9,轮到T也想拿3,6,9这时需要数总共多少个3,6,9总共奇数个则S赢
对于2.S必须拿一个非3,6,9而且使剩下的凑成3的倍数。这时T面临的情况就和S在1.开始时相同,而S若想赢,3,6,9的总和就得是偶数
11489 Integer GameTwo players, S and T, are playing a game where they make alternate moves. S plays first.In this game, they start with an integer N. In each move, a player removes one digit from theinteger and passes the resulting number to the other player.
The game continues in this fashion untila player finds he/she has no digit to remove when that player is declared as the loser.With this restriction, its obvious that if the number of digits in N is odd then S wins otherwise Twins. To make the game more interesting,
we apply one additional constraint. A player can remove aparticular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left.Suppose N = 1234. S has 4 possible moves. That is, he can remove 1, 2, 3, or 4. Of these,
two ofthem are valid moves.• Removal of 4 results in 123 and the sum of digits = 1 + 2 + 3 = 6; 6 is a multiple of 3.• Removal of 1 results in 234 and the sum of digits = 2 + 3 + 4 = 9; 9 is a multiple of 3.The other two moves are invalid.If both players play
perfectly, who wins?InputThe first line of input is an integer T (T < 60) that determines the number of test cases. Each case isa line that contains a positive integer N. N has at most 1000 digits and does not contain any zeros.OutputFor each case, output
the case number starting from 1. If S wins then output ‘S’ otherwise output ‘T’.Sample Input3433771Sample OutputCase 1: SCase 2: TCase 3: T
#include <iostream> #include<cstdio> #include<cstring> using namespace std; int t,m[10],sum,sum1,cnt,temp; char num[1005]; int main() { // freopen("cin.txt","r",stdin); scanf("%d",&t); int cnt=1; while(t--) { printf("Case %d: ",cnt++); scanf("%s",num); memset(m,0,sizeof(m)); sum=0,sum1=0; int len=strlen(num); if(len==1)//一开始这里写成了len=1 ->_-> 能不能用点心 { printf("S\n");continue; } for(int i=0;i<len;i++) { temp=num[i]-'0'; m[temp]++; sum+=(temp); if(temp%3==0) sum1++; } char qq='T'; if(sum%3==0) { if(sum1%2) { qq='S'; } } else //if(sum1%2) { for(int i=0;i<=9;i++) { if(m[i]&&(sum-i)%3==0&&(i%3)) { if(sum1%2==0) qq='S'; break; } } } printf("%c\n",qq); } return 0; }
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