游隼烈天的个人资料

标准版玩家体验感受新版本玩了两天 1，画面总感觉跟手机似的，溜冰 2，人员，传球等特别奇怪，动画感很强，就是球判定到你这了，然后播放动画，然后传球也离谱，一些变扭的传球，会忽然传很远 3，感觉球场变小了，边线离禁区太近了，全是人 4，十字键，忒不习惯总结这一代的变化跟前几代差距非常大，但是改的又不好，不知道ea怎么想的

就问问王异敢不敢削流浪军 50人流浪军能打一个盟，还爽的一批本来是被打败的盟，结果变身流浪军能直接打趴正规军凭什么！强度差距太大，不仅暗改，还无限流正规军拉预备役，还得先训练，要大要塞，还要花体力你流浪军随随便便拉着完别说容易饿死，你一个盟分离出去的，都是喂的饱饱的真的离谱，现在2000认得大区，对面有一个盟上山就能火烧洛阳，况且征服奖励少，高红都无所谓，上山还能打的爽

顶配你们都会，那乞丐阵容有人会搭么刚入手求阵容，目前有个金卡凯恩、卢卡库、苏亚雷斯，求个2w金币以内的阵容，化学大师来畅所欲言吧

随便聊聊主播配将的事同一个账号找了三个主播配，结果出了三套不同的阵容直接蒙圈了，这一次80钱也太好挣了

时隔多年重新玩，各位大佬们1万能买个什么号？时隔多年重新玩，各位大佬们1万能买个什么号？

请问下怎么看两个球员谁好？基耶萨比阿塔尔好嘛？看测评都是说手感的，没队友也无法试验还有化学一定要满分才最好吗，姆巴佩能替补上吗？

这游戏对新手太难了，全是挫败感入坑一个多月，某宝买金凑了套意甲套目前已经掉到d10，100分，爬也爬不上去打电脑好歹也是高分世界级，低分再高一档难度，但是打人真的难受，老是被一脚直塞打爆，高速跟跑还被抗开怒啊，这游戏难度就没有新人了嘛！！！附上阵容

搞疯了，这个任务就是完成不了，求助！求助，传中助攻，我试过直塞助攻，两下直塞助攻，传球，角球头球，结果还是一次都没有。

请问怎么举报别人用外挂？请问怎么举报别人用外挂？

日防夜防家贼难防今天上线家里竟然刷了只33的毒狗，真是日了狗了，毒狗还啃死2个穿铁甲的，2个冰箱，一个大箱子，1个小箱子，2个研磨器。损失2套铁甲，一大堆材料，铁器等。门跟墙壁都是封的严严实实，一半铁一半石头。

请问下怎么拆光别人的遗址好不容易找块理想的居住之地，奈何地上留了几根别人的fence跟柱子，还是石头的，请问该怎么去处理掉这些呢？

关于游戏语言 hold不住拉，入手了小v，不过发现中文版的游戏没几个，全是日文的，这让我等日文文盲如何是好啊。

悲剧啊，前几天还有2个粉丝，今天就1位了骂了隔壁的我还回家过几位

有黑客在不，求黑个网站 http://tieba.baidu.com/mo/q/checkurl?url=http%3A%2F%2Fzhinazhu.forums-free.com%2Ftopic-f1.html&urlrefer=1afd8df950f0e3860f1c424a78adcc56 看网址狗(百度)日的卖(百度)国贼

一直在关注会员申请，结果发现会员人数在减少 RT

KLZ速成班刷帖中，怎么还不能申请啊 KLZ速成班刷帖中，请勿打扰~~~ KLZ速成班刷帖中，请勿打扰~~~ KLZ速成班刷帖中，请勿打扰~~~

KLZ速成班 KLZ速成班刷帖中，请勿打扰~~~ KLZ速成班刷帖中，请勿打扰~~~ KLZ速成班刷帖中，请勿打扰~~~

最后一帖啦，吧主好人别封我啊，好不容易翻出来的账号最后一帖啦，吧主好人别封我啊，好不容易翻出来的账号最后一帖啦，吧主好人别封我啊，好不容易翻出来的账号

就差2帖，吧主好人别封我啊，好不容易翻出来的账号就差2帖，吧主好人别封我啊，好不容易翻出来的账号就差2帖，吧主好人别封我啊，好不容易翻出来的账号就差2帖，吧主好人别封我啊，好不容易翻出来的账号

大家虎年快乐凑10帖，不用回复了

凑10个帖子，另求问怎么看发帖数？怎么查看自己发了几个帖子啊？

怎么还没到达申请会员的要求？怎么还没到达申请会员的要求？怎么还没到达申请会员的要求？

怎么不能申请会员？回帖不算数量？？回帖不算数量？？回帖不算数量？？回帖不算数量？？

祝各位吧主虎年快乐祝各位吧主虎年快乐祝各位吧主虎年快乐祝各位吧主虎年快乐祝各位吧主虎年快乐为什么不能申请会员那？回帖不算？

难道回帖不算发帖数？为什么还不能申请会员啊？？回帖不算发帖数？

问个问题（标题短短的）发帖数10个以上包括回复吗？

为什么不能申请会员啊？？不是说发帖10个，然后注册时间长于7天就OK拉？我好不容易才把这个账号给翻出来的，咋不让申请呢？

求教道题目实在看不出了，刚做搜索题就碰壁了，看了一下午没结果，写的程序是和例子对的上的，可就是一直WA，感觉也看不出来，求高手帮忙看下哪里错啦 http://tieba.baidu.com/mo/q/checkurl?url=http%3A%2F%2F210.32.0.220%2Fonlinejudge%2FshowProblem.do%3FproblemCode%3D1711&urlrefer=55e571a485d7031954a25ef74cde35ac 这是题目的网址，浙大的1711，sum it up #include<stdio.h> #define MAX 1000 int num,sum,total,n,s[MAX],i_path[MAX],path[MAX]; int flag=0,state=0; int compare() { int j,k; j=k=0; while(i_path[j]!='\0'&&path[k]!='\0') { if(i_path[j]!=path[k]) return 0; j++; k++; } if(i_path[j]!=path[k]) return 0; return 1; } void copy() { int j; for(j=0;path[j]!='\0';j++) i_path[j]=path[j]; i_path[j]='\0'; return ; } void dfs(int i) { int j; int k; if(total==sum) { path[num]='\0'; if(!flag||!compare()) { state=1; flag=1; copy(); for(k=0;k<num;k++) { if(!k) printf("%d",path[k]); else printf("+%d",path[k]); } printf("\n"); } return ; } for(j=i+1;j<n;j++) { if(total+s[j]<=sum) { total+=s[j]; path[num++]=s[j]; dfs(j); total-=s[j]; num--; } } return ; } int main() { int l; while(scanf("%d%d",&sum,&n)&&sum) { state=0; flag=0; total=0; num=0; for(l=0;l<n;l++) scanf("%d",&s[l]); printf("Sums of %d:\n",sum); dfs(-1); if(!state) printf("NONE\n"); } return 0; }

求教ZOJ 1082 一道简单的最短路径问题，可就是一直WA，题目是 http://tieba.baidu.com/mo/q/checkurl?url=http%3A%2F%2F210.32.0.220%2Fonlinejudge%2FshowProblem.do%3FproblemId%3D82&urlrefer=1f3487ef60fcfcb54f202bcda6a29cee 下面是我写的一段程序结果与例题的结果一样，求教高手说下程序哪里错了，还是思想就错了 #include<stdio.h> #define MAX 100000 int g[256][256],q[256][256],max[256]; int n,re; int alg() { int ma=MAX; int i,j,k,x,m; re=0; for(i=1;i<=n;i++) { if(g[i][0]>=MAX) break; for(x=1;x<=n;x++) { for(m=1;m<=n;m++) { if(i!=m) { if(g[i][m]>=MAX&&g[g[i][0]][m]<MAX) g[i][m]=g[g[i][0]][m]+g[i][g[i][0]]; else if(g[i][m]<MAX&&g[g[i][0]][m]<MAX) { if(g[g[i][0]][m]+g[i][g[i][0]]<g[i][m]) g[i][m]=g[g[i][0]][m]+g[i][g[i][0]]; } } } q[i][g[i][0]]=1; for(k=1;k<=n;k++) if((g[i][k]<g[i][g[i][0]]||q[i][g[i][0]])&&!q[i][k]&&i!=k) g[i][0]=k; if(q[i][g[i][0]]) break; } for(j=1;j<=n;j++) { if(i!=j) { if((g[i][j]>max[i]||max[i]>=MAX)&&g[i][j]<MAX) max[i]=g[i][j]; if(g[i][j]>=MAX) { max[i]=MAX; break; } } } } for(i=1;i<=n;i++) if((max[i]<ma)&&max[i]<MAX) { ma=max[i]; re=i; } return ma; } int main() { int o,w,x,c; int ma; int i,j; while(scanf("%d",&n)==1&&n) { for(i=0;i<256;i++) { max[i]=MAX; for(j=0;j<256;j++) { g[i][j]=MAX; q[i][j]=0; } } for(i=1;i<=n;i++) { scanf("%d",&o); for(w=1;w<=o;w++) { scanf("%d%d",&x,&c); g[i][x]=c; } } for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if((g[i][j]<g[i][0]||g[i][0]>=MAX)&&g[i][j]<MAX&&i!=j) g[i][0]=j; } } ma=alg(); if(n==1) printf("%d 0\n",n); else if(ma<MAX) printf("%d %d\n",re,ma); else printf("disjoint\n"); } }

我刚在百度知道里提问了一个问题，怎么就没了啊？ RT我提问了一个C#编程方面的问题，结果我点我的问题里就找不到了我还为这个问题设置了20分的啊？？？

请教下900法伤的纯冰法能下什么YXFB 请教下900法伤的纯冰法能下什么YXFB,现在都只下个YXCQ,NY,ST的，怕法伤低了拖累别人FB意识还可以，羊怪补羊，帮治疗冰环怪，OT的时候闪现+冰箱都没出过什么错误请教达人法师，我是点的61冰系

求个c语言动态规划求最短路径的题目以及源代码 RT求个c语言动态规划求最短路径的题目以及源代码希望有的能够贴下，谢谢

请教道题目 http://acm.zju.edu.cn/show_problem.php?pid=1905Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n). InputEach test case is a line of input representing s, a string of printable characters. OutputFor each s you should print the largest n such that s = a^n for some string a. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case. Sample Inputabcdaaaaababab.Sample Output143

浙大ACM1205 http://acm.zju.edu.cn/show_problem.php?pid=1205In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest. As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers. Input:You're given several pairs of Martian numbers, each number on a line. Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19). The length of the given number is never greater than 100.Output:For each pair of numbers, write the sum of the 2 numbers in a single line.Sample Input:1234567890abcdefghij99999jjjjj9999900001Sample Output:bdfi02467jiiiij00000

求角改下 The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.InputThere are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.OutputPrint the total time on a single line for each test case. Sample Input1 23 2 3 10Sample Output1741这是题目

请教个问题，在线 #include int main(){int n;char c;scanf("%d",&n);while(n--){while(c=getchar()!=EOF){if(c=='Z')c='A';else c=1+c;printf("%c",c);}}return 0;}为什么输入n的值后最输出一个不知道什么的东东啊？？？

浙大ACM1037 BackgroundThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.InputThe input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.OutputFor each integer in the input, output its digital root on a separate line of the output.ExampleInput24390Output63

很懵很懵 ProblemThe president of Gridland has hired you to design a program that calculates the length of the shortest traveling-salesman tour for the towns in the country. In Gridland, there is one town at each of the points of a rectangular grid. Roads run from every town in the directions North, Northwest, West, Southwest, South, Southeast, East, and Northeast, provided that there is a neighbouring town in that direction. The distance between neighbouring towns in directions North–South or East–West is 1 unit. The length of the roads is measured by the Euclidean distance. For example, Figure 7 shows 2 × 3-Gridland, i.e., a rectangular grid of dimensions 2 by 3. In 2 × 3-Gridland, the shortest tour has length 6. Figure 7: A traveling-salesman tour in 2 × 3-Gridland.InputThe first line contains the number of scenarios.For each scenario, the grid dimensions m and n will be given as two integer numbers in a single line, separated by a single blank, satisfying 1 < m < 50 and 1 < n < 50.OutputThe output for each scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. In the next line, print the length of the shortest traveling-salesman tour rounded to two decimal digits. The output for every scenario ends with a blank line.Sample Input22 22 3Sample OutputScenario #1:4.00Scenario #2:6.00 这是题目,我看了N久N久才明白点,然后写了#include #include main(){int i,n,h,l;float sum;scanf("%d",&n);for(i=1;i<=n;i++){ scanf("%d%d",&h,&l); if(h%2==0)sum=h*l-1+h-1;else sum=h*l-1+(sqrt(2))*(l>h?(h-1):(l-1))+fabs(l-h);printf("Scenario #%d:\n",i);printf("%.2f\n",sum);printf("\n");}}改了N次N次还是错,没辙了,求救现在才懂得英语的重要性啊

求教道题目题目：一个整数，它加上100后是一个完全平方数，再加上168又是一个完全平方数，请问该数是多少？#include main(){int n,i,loop;for(n=1;n<=10000;n++){loop=0;for(i=1;i<+100;i++)switch(loop){case 0 :if(n==i*i) loop+=1;break;case 1: if(n+100==i*i)loop+=1;break;case 2: if(n+268==i*i) loop+=1;break;}if(loop==3) printf("%d",n);}}我知道这样写很愚蠢......但就想知道错在哪里麻烦知道的说下,谢谢

求教个问题作为FLOAT型变量,计算的时候是先转化成double型的,那么sizeof(x+y)这个该是多少???定义x和y为float型的

请教个共用体方面的小问题 union u{ int k;char c[2];}u1; main() { u1.c[0]=13; u1.c[1]=0; printf("%d\n",u1.k); }解释上说k占有2个字节，这2字节分别被赋值为0和13，那为什么只输出13？？而不是013或则130？

求教c语言递归调用啊!! 本人是自学的c语言.看到前面靠着习题跟书本都还可以,就是到了递归调用地方就不懂了,不明白递归调用是怎么运行的,还有找不出与数学知识的连接,望高手帮忙解释下啊,越全面越好万分感谢

求教c语言递归调用啊!!

养由翎怎么加入啊?? 我第一章都快结束好久了,在地图上乱跑，就是没发动养由翎的任务,悬赏养由弓这个任务我没碰到过,不知道出发这个任务需要什么条件?听说这MM弓箭不错,不知道是否值得练

【甄爱馥现】谁有HEBE的生活照啊?? 谁有HEBE的生活照啊????上次看到几张,觉得比银幕上的漂亮多了