qq_35785913的博客

#include <iostream>#include<cstdio> #include <algorithm>using namespace std;int main() { int first,n,k,temp; scanf("%d%d%d",&first,&n,&k); int data[100010...

#include<cstdio>int main(){ int count[10]; for(int i=0;i<10;i++) { scanf("%d",&count[i]); } for(int i=1;i<10;i++) { if(count[i]!=0) { printf("%d",i); count[i]--; ...

真的很需要细心！！#include<cstdio>#include<algorithm>using namespace std;bool cmp(int a,int b){ return a>b;}int main(){ int n,m,a[110]; scanf("%d",&n); bool hash[1000]={0}; for(i...

#include<cstdio>const int maxn=1010;int hash[maxn]={0};int main(){ int n; scanf("%d",&n); int team,member,score; for(int i=0;i<n;i++) { scanf("%d-%d %d",&team,&member,&...

#include<cstdio>const int maxn=10010;char dist[6]={'P','A','T','e','s','t'};int hash[6]={0};int main(){ char str[maxn]; gets(str); int len=strlen(str); int sum=0; for(int i=0;i<len;...

#include<cstdio>#include <cctype>#include <string>using namespace std;int main() { string s; getline(cin, s); int a[26] = {0}; for (int i = 0; i < s.length(); ...

#include<cstdio>int change(char c)///此处就相当于定义了散列函数！！ { if(c>='0'&&c<='9') return c-'0'; if(c>='a'&&c<='z') return c-'a'+10; if(c>='A'&&c<='Z') retu

用空间换时间，hash数组的思想，原来正式名称叫散列啊#include<cstdio>int main(){ int n,score; int hash[100]={0}; scanf("%d\n",&n); for(int i=0;i<n;i++) { scanf("%d",&score); hash[score]++; } int k...

#include <iostream>#include <string>#include<cctype> using namespace std;int main() { string bad, should; getline(cin, bad); //为了防止第一行是空的，不能用cin >> ,用getline(cin, ...)...

#include <iostream>#include<string>using namespace std;int main() { string s1, s2, ans; cin >> s1 >> s2; for (int i = 0; i < s1.length(); i++) if (s2....

#include<cstdio>#include<vector>#include<algorithm>using namespace std;struct mooncake{ double store,sell,price; };bool cmp(mooncake a,mooncake b){ return a.price>b.pric...

结论：能使选出的数的个数最大的方案，一定是在该递增序列中选择连续的若干个数的方案。问题转换成：在一个给定的递增序列中，确定一个左端点a[i]和右端点a[j],且j-i的值最大#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;bool cmp(int a,int b...

#include <iostream>#include<cstdio>#include <string>using namespace std;void func1(string s);void func2(string s);int main() { int n; scanf("%d",&n); getchar...

#include<cstdio>#include<cmath>const int maxn=1000000000;bool isPrime(int n){ if(n<=1) return false; int sqr=(int)sqrt(1.0*n); for(int i=2;i<=sqr;i++) { if(n%i==0) return f...

#include<cstdio>#include<cmath>bool isPrime(int n){ if(n<=1) return false; int sqr=(int)sqrt(1.0*n); for(int i=2;i<=sqr;i++) { if(n%i==0) return false; } return true;}in...

#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;struct Fraction{ ll up,down;}a,b;ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b);}Fraction reduction(...

思路：模拟手动除法过程，每一位够除的写到最终商的结果中，余数*10加后一位的值继续进行除法操作。#include <iostream>#include<string>using namespace std;int main() { string s; int a, t = 0, temp = 0; cin >> s >>...

规律：如果当前是第i个数，那么其出现总次数等于i*(n+1-i)#include<cstdio>int main(){ int n; double v,ans=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%lf",&v); ans+=v*i*(n+1-i); } printf("%.2...

#include<cstdio>#include<algorithm>using namespace std;bool cmp1(int a,int b){return a>b;}bool cmp2(int a,int b){return a<b;}void chai(int n,int num[]){ for(int i=0;i<4;i...

总结字符串要求：只能有一个P一个T，中间末尾和开头可以随便插入A。但是必须满足开头的A的个数 * 中间的A的个数 = 结尾的A的个数#include <iostream>#include<cstdio>#include<string>#include <map>using namespace std;int main() { int...

暴力判断会超时，考虑继承关系。令数组leftmax记录序列A的每一位左边的最大值，不含本位令数组rightmin记录序列A的每一位右边的最小值，不含本位很棒的思想！#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;const int INF=1000000000;...

思路：对一个确定位置的A来说，以它形成的PAT的个数等于它左边p的个数乘以它右边T的个数还有右侧不用定义数组，边统计边计算最后结果。#include<cstdio>#include<cstring>const int maxn=100010;const int mod=1000000007;int main(){ char leftp[maxn]={0}; c...

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct student{ char id[10]; int de,cai,sum; int flag;}stu[100010];bool cmp(student a,student b){ ...

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=110;char a[maxn],b[maxn];char ans[maxn]={0};int main(){ scanf("%s %s",a,b); int lenA=...

#include<cstdio>struct Student{ char name[10]; char id[15]; int score;}temp,max,min;int main(){ int n; max.score=-1; min.score=101; scanf("%d",&n); for(int i=0;i<n;i++) { ...

#include<cstdio> const int maxn=1000; struct Student { long long number; int ty; int te; }student[maxn]; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { ...

#include<cstdio>int main(){ int m,n; int flag=0; while(scanf("%d %d",&m,&n)!=EOF) { if(flag==0&&n==0) printf("0 0"); if(n!=0) { if(flag==1) printf(" "); printf(..

思路：一种一种情况做判断，实在是太麻烦 #include<cstdio>int main(){ int n; scanf("%d",&n); int timesA[3]={0},timesB[3]={0}; int handA[3]={0},handB[3]={0};//BCJ for(int i=0;i<n;i++) { char c1,c2;...

#include<cstdio>int count[5]={0};int ans[5]={0};int main(){ int n; scanf("%d",&n); int temp; for(int i=0;i<n;i++) { scanf("%d",&temp); if(temp%5==0&&temp%2==0) ...

思路：定义交换函数reverse 0......(n-m-1),(n-m),........(n-1)                             1换：（n-m-1）......0,(n-m),........(n-1)                              2换：(n-m-1)..........0,(n-1).............(n-m)        ...

#include<cstdio>int main(){ int N; scanf("%d",&N); int counta=0,countb=0; int a,b,c,d; for(int i=0;i<N;i++) { scanf("%d%d%d%d",&a,&b,&c,&d); if(a+c==b&&amp

思路：已知c1,c2两个时钟打点数，除100即得到总秒数/3600得到小时数，%3600/60得到分钟数，%3600%60得到秒#include<cstdio>int main(){ int c1,c2; scanf("%d%d",&c1,&c2); int n=(c2-c1+50)/100; int hour=n/3600; int minute=n%3...

思路：主要实现在整数a中查找是否存在某一位的值为da由于10^10超过了int型的范围，定义为long long型枚举a的每一位，定义pa用来保存a中存在的da #include<cstdio> int main() { long long a,b,da,db; scanf("%lld%lld%lld%lld",&a,&da,&b,&db);...

#include <cstdio>int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { long long int a, b, c; scanf("%lld%lld%lld", &a, &b, &c); ...

#include<cstdio>struct person{ char name[10]; int year; int month; int day;}max,min;int main(){ int n; max.year = 0, max.month = 0, max.day = 0, min.year = 9999, min.month = 9999, ...

#include<cstdio>int main(){ int n; scanf("%d",&n); int num,score; int c[100000]={0}; for(int i=0;i<n;i++) { scanf("%d %d",&num,&score); c[num]+=score; } int max=0; ...

（unfinished）自己写了一半，后面写不出了，参考书上的也看不懂了，郁闷~#include<cstdio>int main(){ char str[10010]; gets(str); int len=strlen(str); if(str[0]=='-') printf("-"); int pos=0; for(int i=0;i<len;i++) {...

注意：“%02d”每种判别之间的转换需要理清楚#include<cstdio>#include<cstring>int main(){ char week[7][5]={"MON","TUE","WED","THU","FRI","SAT","SUN"}; char str1[70],str2[70],s

思路：以空格为分隔符统计字符个数，将单词顺序保存在二维数组中，最后逆序输出。输入完内容先回车再输入ctrl+z再回车确认文件结束Scanf（）函数有一个返回值，成功时返回的是成功输入数据的个数，错误时返回EOF,EOF为宏定义的一个值#include<cstdio>int main(){ int num=0; char str[90][90]; while(scanf("%s...

int main(){ char str[100]; scanf("%s",str); int len=strlen(str); int sum=0; for(int i=0;i<len;i++) { sum+=str[i]-'0'; //*** } int ans[10]; int count=0; do { ans[count++]=sum%10; ...