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RSS订阅原创 POJ2083 ZOJ2423 Fractal
//3_4_1: Fractal 分形打印X POJ2083 ZOJ2423#include #include #include using namespace std;char map[730][730];void print(int n,int x,int y){ int i,m = 1; if(n == 1) { map[x][y] = 'X'; return;
2015-04-28 11:32:05
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原创 POJ2013 ZOJ2172 UVA 3055Symmetric Order
//3_3_1: Symmetric Order 姓名重新排列输出 POJ2013 ZOJ2172 UVA 3055#include #include #include using namespace std;char str[16][30];int main(){ int i,n,cnt = 1; while(scanf("%d",&n),n != 0) { for(i
2015-04-27 20:45:38
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原创 POJ1979 ZOJ2165 Red and Black
//3_2_1: Red and Black 到达黑色瓷砖数量 POJ1979 ZOJ2165#include #include #include using namespace std;int i,j,mx,my,sx,sy,cnt,dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};char enter,map[22][22];void DFS(int
2015-04-27 19:46:27
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原创 POJ1281 ZOJ1371 UVA2514 MANAGER
//2_4_5: MANAGER 进程管理的模拟 POJ1281 ZOJ1371 UVA2514#include #include #include using namespace std;const int maxn = 10000 + 10;int spend[maxn],delet[maxn],queue[maxn];void quick_sort_max(int st,int
2015-04-27 17:14:41
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原创 POJ2864 UVA3470 Pascal Library
//2_4_3: Pascal Library 是否有人参加全部晚宴 POJ2864 UVA3470#include #include #include using namespace std;int main(){ int N,D,i,j,temp,flag,a[100 + 10]; while(1) { scanf("%d%d",&N,&D); if(N == 0 |
2015-04-26 14:23:20
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原创 POJ1207 UVA100 The 3n+1 problem
//2_4_2: The 3n+1 problem NP问题 POJ1207 UVA100#include #include #include using namespace std;int main(){ int a,b,i,j,min,max,ans,time; while(scanf("%d %d",&a,&b) != EOF) { if(a > b) { m
2015-04-26 13:53:52
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原创 POJ2000 ZOJ2345 UVA3045 Gold Coins
//2_4_1: Gold Coins 国王赏赐金币问题 POJ2000 ZOJ2345 UVA3045#include #include #include using namespace std;int main(){ int i,N,sum,day; while(scanf("%d",&N), N != 0) { day = sum = 0; for(i = 1;;i
2015-04-26 11:38:09
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原创 UVA11000 Bee
//2_3_1: Bee 非洲雌蜂和雄蜂繁殖问题 UVA11000#include #include #include using namespace std;int main(){ unsigned a,b,t; int N,i; while(scanf("%d",&N),N != -1) { a = 0; b = 1; for(i = 0;i < N;i ++)
2015-04-26 11:03:16
203
原创 POJ1316 ZOJ1180 UVA640 Self Numbers
//2_2_1: Self Numbers 10000以内的自数 POJ1316 ZOJ1180 UVA640#include #include #include using namespace std;const int N = 10000 + 100;int sum(int n){ if(n < 10) return n; else return (n % 10) + su
2015-04-26 00:22:36
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原创 POJ1922 ZOJ2229 Ride to School
//2_1_2: Ride to School 骑自行车跟上同学 POJ1922 ZOJ2229#include #include #include using namespace std;const double limit = 100000000;int main(){ int i,N,time,speed; double min,sum; while(scanf("%d"
2015-04-25 23:00:54
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原创 POJ2017 ZOJ2176 UVA3059 Speed Limit
//2_1_1: Speed Limit 里程计算 POJ2017 ZOJ2176 UVA3059#include #include #include using namespace std;int main() { int i,n,sum,time,temp,speed; while(scanf("%d",&n), n != -1) { sum = time = 0;
2015-04-25 19:42:59
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原创 POJ1019 ZOJ1410 Number Sequence
//1_5_10: Number Sequence 求连续数组成的数组串的某位数 POJ1019 ZOJ1410#include #include #include #include using namespace std;unsigned a[31270],s[31270];void init(){ int i; a[1] = s[1] = 1; for(i = 2;i <
2015-04-25 16:00:49
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原创 POJ1833 排列
//1_5_9: 排列 求下k个排列组合 POJ1833#include #include #include using namespace std;int a[1030];void quick_sort(int st,int ed) //快排算法{ int i = st,j = ed,temp; if(st < ed) { temp = a[st]; while(
2015-04-25 13:51:33
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原创 POJ1338 Ugly Numbers
//1_5_8: Ugly Numbers 丑陋数 POJ1338#include #include #include using namespace std;int min(int a,int b,int c){ int temp = a < b ? a : b; return temp < c ? temp : c;}int main(){ int i,l,m,n,cn
2015-04-24 17:02:17
173
原创 POJ2136 Vertical Histogram
//1_5_7: Vertical Histogram 统计字符个数并输出 POJ2136#include #include #include using namespace std;int main(){ int i,j,max,let[26] = {0}; char str[80],letter; for(i = 0;i < 4;i ++) { gets(str);
2015-04-23 21:00:59
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原创 POJ2242 ZOJ1090 The Circumference of the Circle
//1_5_6: The Circumference of the Circle 三点确定圆 POJ2242 ZOJ1090#include #include #include #include using namespace std;const double pi = 3.141592653589793;int main(){ double x1,y1,x2,y2,x3,y3,
2015-04-23 20:32:07
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原创 POJ3006 Dirchlet's Theorem on Arithmetic Progressions
//1_5_5: Dirchlet's Theorem on Arithmetic Progressions 写第i个素数 POJ3006#include #include #include #include using namespace std;int isprime(int n){ int i; int sq = (int)sqrt((double)n) + 1; if(
2015-04-23 18:05:16
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原创 POJ1581 ZOJ1764 A Contesting Decision
//1_5_4: A Contesting Decision 判定竞赛题目与罚时 POJ1581 ZOJ1764#include #include #include using namespace std;struct node{ char name[20]; int cnt,pun;}team[100];int main(){ int n,i,j,num,time,max
2015-04-23 17:16:16
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原创 POJ3094 ZOJ2812 Quicksum
//1_5_3: Quicksum 求字符串的校验 POJ3094 ZOJ2812#include #include #include using namespace std;int main(){ int i,sum; char str[260]; while(gets(str),str[0] != '#') { i = sum = 0; while(str[i])
2015-04-23 16:22:02
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原创 POJ2196 ZOJ2405 Specialized Four_Digit Numbers
//1_5_2: Specialized Four_Digit Numbers 10,12和16进制的数字和相等 POJ2196 ZOJ2405#include #include #include using namespace std;int main(){ int i,j,sum1,sum2,sum3; for(i = 2992;i < 10000;i ++) { sum
2015-04-23 16:05:55
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原创 Ural1068 Sum
//1_5_1: Sum 求1到N之间的整数和 Ural1068#include #include #include using namespace std;int main(){ int N; scanf("%d",&N); if(N > 0) printf("%d\n",(1 + N) * N / 2); else printf("%d\n",(1 - N) * N /
2015-04-23 15:34:59
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原创 POJ3299 Humidex
//1_4_2: Humidex 湿热指数，温度和露点之间的关系 POJ3299#include #include #include #include using namespace std;double get_hum(double tem,double dew){ return tem + 0.5555 * (6.11 * exp(5417.7530 * ((1 / 273.1
2015-04-22 22:50:20
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原创 POJ1003 ZOJ1045 Hangover
//1_4_1: Hangover 卡片伸向桌子外 POJ1003 ZOJ1045#include #include #include using namespace std;const int maxn = 300;double len[maxn];int main(){ int i,n = 1; double length; len[n] = 0.5; while(le
2015-04-22 22:05:45
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原创 POJ1005 ZOJ1049 I think I need a houseboat
//1_3_1: I think I need a houseboat 河水淹没土地 POJ1005 ZOJ1049#include #include #include using namespace std;const double pi = 3.1415926;int main(){ int tem,N,i = 1; double x,y,ans; scanf("%d",&
2015-04-22 19:21:00
161
原创 POJ2739 Sum of Consecutive Prime Numbers
//1_2_2: Sum of Consecutive Prime Numbers 连续素数之和 POJ2739#include #include #include #include using namespace std;const int maxn = 2000;int isprime(int n){ int i; if(n == 2) return 1; int sq
2015-04-22 17:48:02
181
原创 POJ1552 ZOJ1760 Doubles
//1_2_1: Doubles 求2倍的个数 POJ1552 ZOJ1760#include #include #include using namespace std;int main(){ int i,j,n,cnt,a[20]; scanf("%d",&a[0]); while(a[0] != -1) { n = 1; while(scanf("%d",&a[n
2015-04-22 16:43:19
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原创 POJ1004 ZOJ1048 Financial Management
//1_1_1: Financial Management 月收入平均值 POJ1004 ZOJ1048#include #include #include using namespace std;int main(){ double ave,sum = 0,mon; int i; for(i = 0;i < 12;i ++) { scanf("%lf",&mon); //
2015-04-22 16:12:33
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原创 POJ1035 ZOJ2040 Spell checker
//6_3_1: Spell checker 检查最相近的单词并输出 POJ1035 ZOJ2040#include #include #include using namespace std;char word[10000][16];int match(char str1[],int len,char str2[]){ int i,pos1,pos2,flag = 0; for
2015-04-19 20:39:01
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原创 POJ2503 Babelfish
//6_1_2: Babelfish 英文单词和外语单词之间的转换 POJ2503#include #include #include using namespace std;const int maxc = 10 + 5;const int maxn = 100000 + 5;int n;struct info{ char org[maxc]; char trs[maxc]
2015-04-18 23:45:48
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原创 POJ1706 References
//6_1_1: References 参考文献的整理 POJ1706#include #include #include using namespace std;const int maxl = 80 + 5; //一行最长为80个字符const int maxs = 240 + 10; //参考文献不超过3行const int maxr = 1000; //参考文献最多设置10
2015-04-18 14:49:49
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原创 ZOJ1004 Anagrams by Stack
//5_4_5: Anagrams by Stack 通过一系列栈操作将字符翻转 ZOJ1004#include #include #include using namespace std;int i,len,top,str_len,dst_len;char str[100],dst[100],ans[100],stack[100];void dfs(int x,int y){
2015-04-16 17:17:50
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原创 POJ3044 City Skyline
//5_4_4: City Skyline 根据楼的叠加效果求楼的数量 POJ3044#include #include #include using namespace std;const int maxn = 50000 + 10;const int maxh = 25000 + 10;int main(){ int i,x,N,W,top = 0,cnt = 0,a[max
2015-04-16 10:43:24
203
原创 POJ1012 Joseph
//5_4_3: Joseph k个好人和坏人判断步长问题 POJ1012#include #include #include using namespace std;int main(){ int i,k,m,n,s,t,flag,ans[15]; for(k = 1;k <= 13;k ++) { i = 1; n = 2 * k; s = t = flag =
2015-04-15 20:32:55
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原创 POJ1591 ZOJ1326 M*A*S*H
//5_4_2: M*A*S*H 通过卡片数字去掉人员 POJ1591 ZOJ1326#include #include #include using namespace std;int main(){ int i,j,k,N,X,cnt,max,a[25],tag[55],loop = 1; while(scanf("%d",&N) != EOF) { scanf("%d"
2015-04-14 15:07:33
367
原创 POJ3125 Printer Queue
//5_3_3: Printer Queue 打印机优先队列问题 POJ3125#include #include #include using namespace std;const int maxn = 100 + 5;int main(){ int i,n,m,k,a[maxn],cnt,tag,head,rear,loop,print; scanf("%d",&loop)
2015-04-13 23:43:49
165
原创 POJ2259 ZOJ1948 Team Queue
//5_3_2: Team Queue 团队队列的基本操作 POJ2259 ZOJ1948#include #include #include using namespace std;const int maxn = 1000000 + 10; //总的元素个数const int maxb = 200000 + 10; //总的子队列数const int maxs = 1000
2015-04-13 17:23:55
232
原创 ZOJ3210 A Stack or A Queue
//5_3_1: A Stack or A Queue 判断是栈还是队列 ZOJ3210#include #include #include using namespace std;const int maxn = 100 + 10;int main(){ int i,x,N,loop,a[maxn],isstack,isqueue; cin >> loop; while(lo
2015-04-12 16:14:52
225
原创 POJ2106 ZOJ2483 Boolean Expressions
//5_2_2: Boolean Expressions 布尔表达式求值 POJ2106 ZOJ2483#include #include #include using namespace std;/***********优先级比较***********//*-----------------------------*//* '(' | '|' | '&' | '!' | ')' *
2015-04-12 15:57:34
636
原创 POJ1363 ZOJ1259 Rails PopPush
//5_2_1: Rails PopPush城火车站出入组合问题 POJ1363 ZOJ1259#include #include #include using namespace std;const int maxn = 1000 + 10;int main(){ int i,x,N,max,cnt,valid,p[maxn]; cin >> N; while(N) {
2015-04-11 16:47:02
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原创 POJ3750 小孩报数问题
//5_1_1: 小孩报数问题，即约瑟夫问题 POJ3750#include #include #include using namespace std;int main(){ short i,N,w,s,cnt = 1,tag[64] = {0}; char c,name[64][15]; cin >> N; for(i = 0;i > name[i]; cin >> w
2015-04-11 11:13:39
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Test Driven Development of a Django RESTful API
2018-07-31
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