PIANKEzzc的博客

原创 欢迎使用优快云-markdown编辑器

The contest’s message distribution is a big thing in prepare. Assuming N students stand in a row, from the row-head start transmit message, each person can transmit message to behind M personals, and h

原创 欢迎使用优快云-markdown编辑器

#include <iostream>#include<stdio.h>#include <algorithm>#include<cstring>using namespace std;int c[1005], v[1005], f[1005][1005];int main(){ int i, j, n, m, t; cin>>t; while(t--)

原创 文章标题

#include<iostream>#include<algorithm>#include<cmath>using namespace std;int main(){ int i, j, n, m; int sum=0; int a[105], b[105], book[105]={0}; cin>>n; for(i=1; i<=n; i++)

原创 文章标题

#include<iostream>using namespace std;int a[100];int n,ans;int main(){ cin>>n; if (n==0) cout<<10<<endl; else if (n<10) cout<<n<<endl; else { ans=0;//位数数值只有在2-9。

原创 Bessie hears that an extraordinary meteor shower is coming; reports say that these meteors will cras

这是自己关于poj3669写的一部分的代码，一直出错，然后就不想写了。不知道自己哪里错了#include#include#include#includeusing namespace std;int Map[305][305]={0}, m;int dx[5]={0, 0, 0, 1, -1};int dy[5]={0, 1, -1, 0, 0};int

原创 It's the year 4527 and the tanks game that we all know and love still exists. There also exists Grea

It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go i

原创 While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was n

While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only

原创 康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 ♂

康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 ♂舍长有 N 个房间. 第 i 个房间有 J[i] 的 ♂ 需要 F[i] 斤的食物. 康康可以不换完整个房间的♂ ,他可以用 F[i]* a% 的食物 换 J[i]* a% 的 ♂ 现在给你一个实数 M 问你康康最多能获得多少的 ♂InputThe input consists of multiple t

原创 八皇后问题

#include#includeusing namespace std;bool check(int a[], int n);int huanghou(int n);int main(){    int n, m;    cin>>n;    while(n--)    {        cin>>m;        huanghou(m); 

原创 关于将m个苹果放到n个盘子中

此题就像将一个数m分成n个数的加法一样，因为此题不考虑顺序问题，所以我们在分类之前就就将最小的放到第一位。1,如果第一个盘子为0，则接下来m个苹果和n-1个盘子，为app（m, n-1),，当第一个盘子不是0的时候，使所有盘子的个数都加1，即app（m-n, n)这样接下来的的所有分配都离不开这两种情况。最后讨论n==0, m==0, n==1等情况。。。#include#include

原创 Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other.

#include#include#include#include#includeusing namespace std;struct frame{    double x;    double y;}c[1001];struct range{    double l, r;}g[1001];bool cmp(range a

转载 欢迎使用优快云-markdown编辑器

欢迎使用Markdown编辑器写博客本Markdown编辑器使用StackEdit修改而来，用它写博客，将会带来全新的体验哦：Markdown和扩展Markdown简洁的语法代码块高亮图片链接和图片上传LaTex数学公式UML序列图和流程图离线写博客导入导出Markdown文件丰富的快捷键快捷键includeusing namespace std;enum city{Beijing

转载 F

#include#include#include#define NUM 100001using namespace std;int main(){    int n, m, s, sum[NUM], i;    int a[NUM];    scanf("%d", &n);    while(n--)   

转载 A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a p

原创 - Hero

#include#include#includeusing namespace std;struct hero{    double hp;    double dps;}h[21];bool cmp(hero a, hero b){    return a.hp/a.dps }int main(){    int

原创 Crossing River

#include#include#includeusing namespace std;int cmp(const void *a,const void *b){    return *(int *)a-*(int *)b;}int main(){    int t, n, a[1001], i;    scanf("%d", &

原创 今年暑假不AC

#include#includeusing namespace std;struct start{    int s;    int e;}Ti[101];bool cmp(start a, start b){    return a.e}int main(){    int n, i, sum;    while

原创 发工资咯：）

#include#includeint main(){    int n, a[101], sum, i;    memset(a, 0, sizeof(a));    while(~scanf("%d", &n), n != 0)    {        for(i=0; i        {            scanf("%d", &a[i

原创 Saving HDU

#include#includeusing namespace std;struct bag{    int pi;    int mi;}a[105];bool cmp(bag a, bag b){    return a.pi>b.pi;}int main(){    int i, v, n, m, sum;    while

原创 进制转换

#include#includeint main(){    int n, r, i;    int a[1000];    memset(a, 0, sizeof(a));    while(~scanf("%d%d", &n, &r))    {        i = 0;        if(n        {           

原创 olympiad

#include#includeint olinpid(int k);int main(){    int n, a, b, i;    int ans[100001];    memset(ans, 0, sizeof(ans));    for(i=1; i    {        if(olinpid(i))        {     

原创 多个数的最小公倍数

#includeint gongyueshu(int a, int b);int main(){    int n, m, a, b, sum;    while(~scanf("%d", &n))    {        for(int i=0; i        {            sum = 0;            b=1;

原创 鸡兔同笼

#includeint main(){    int i, j, x, y, a;    while(~scanf("%d", &a))    {        if(a%2 != 0)            printf("0 0\n");        else if(a%4 != 0)            printf("%d %d\n", a/4+1,

原创 整数解

#includeint main(){    int i, j, n, m, ture;    while(~scanf("%d%d", &n, &m))    {        if(n==0 && m==0)            break;        ture =0;        for(i=-9999; i        {   

原创 空心三角形

#include#includeint main(){    int i, j, k, n, m=0;    char fl, a[100][100];    while(~scanf("%c", &fl))    {        if(fl == '@')            break;        else            sc

原创 三角形

#includeint a[10];int ture(int a, int b, int c);int main(){    int i, m, a, b, c;    scanf("%d", &m);    for(i=0; i    {        scanf("%d%d%d", &a, &b, &c);        if(ture(a, b

原创 最大公约数

#includeint MAXgongyueshu(int a, int b);int main(){    int a, b;    scanf("%d%d", &a, &b);    printf("a and b MAX_gongyueshu = %d", MAXgongyueshu(a, b));    return 0;}int MAX