0801题解
来源:互联网 发布:二维旋转矩阵的推导 编辑:程序博客网 时间:2024/06/16 18:12
#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;char a[11];char b[11];int main(){scanf("%s %s",&a,&b);int sum1=0,sum2=0;int len1=strlen(a),len2=strlen(b);for(int i=0;i<len1;i++){ sum1+=a[i]-'0';}for(int i=0;i<len2;i++){ sum2+=b[i]-'0';}printf("%d\n",sum1*sum2);return 0;}
#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;int a[5];char s[100005];int main(){for(int i=1;i<=4;i++){ scanf("%d",&a[i]);}scanf("%s",&s);int len=strlen(s);int ans=0;for(int i=0;i<len;i++){ ans+=a[s[i]-'0'];}printf("%d\n",ans);return 0;}
#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;int a[100005];int dp[100005];int main(){int n;while(~scanf("%d",&n)){ if(n==0) break; int sum=0,temp=-1000000009; for(int i=0;i<n;i++){ scanf("%d",&a[i]); sum+=a[i]; if(sum>temp) temp=sum; dp[i]=temp; if(sum<0) sum=0; } int ans=-1000000009; sum=0; for(int i=n-1;i>0;i--){ sum+=a[i]; if(dp[i-1]+sum>ans) ans=dp[i-1]+sum; if(sum<0) sum=0; } printf("%d\n",ans);}return 0;}
#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;struct Node{ int id; int rating; int ans;}r[300005];int cmpr(Node a,Node b){ return a.rating<b.rating;}int cmpd(Node a,Node b){ return a.id<b.id;}int main(){int n;scanf("%d",&n);for(int i=1;i<=n;i++){ scanf("%d",&r[i].rating); r[i].id=i;}sort(r+1,r+n+1,cmpr);int temp=r[1].rating;for(int i=1;i<=n;i++){ if(r[i].rating<=temp){ r[i].ans=temp; temp++; } else{ r[i].ans=r[i].rating; temp=r[i].rating+1; }}sort(r+1,r+n+1,cmpd);for(int i=1;i<n;i++) printf("%d ",r[i].ans); printf("%d\n",r[n].ans);return 0;}
#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>typedef long long ll;using namespace std;bool prime(ll x){if (x%2 == 0) return false; for (ll j = 3; j*j <= x; j += 2) if (x%j == 0) return false; return true;}/*ll gcd(ll a,ll b){ return b==0 ? a : gcd(b,a%b);}*/ll gcd(ll a,ll b){ if(b==0) return a; a%=b; return gcd(b,a);}int main(){int t;scanf("%d",&t);while(t--){ ll n; cin>>n; ll v=n,u=n+1; while(!prime(v)) v--; while(!prime(u)) u++; ll low=v*u*2; ll up=v*u-2*v-2*(u-n-1); ll g=gcd(low,up); up/=g,low/=g; printf("%I64d/",up); printf("%I64d\n",low); //cout << up << "/" << low << endl;}return 0;}
0 0
- 0801题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解
- 题解~~~~
- 题解。。。。
- 题解
- 题解
- 1002 题解
- pku1001题解
- android编译系统makefile(Android.mk)写法
- Java Socket通信及心跳包
- 机房收费系统系列二:MDI子窗体和主窗体显示
- android返回键监听中return false,return true的区别
- 贪心-区域覆盖问题
- 0801题解
- python3.X使用urllib与2.X不同
- Oracle 数据库中批量读取图片
- hdu-4864 Task 2014多校联赛第一场 贪心
- windows下安装jmeter
- linux驱动开发重点关注内容--摘自《嵌入式Linux驱动模板精讲与项目实践》
- asii和unicode格式字符串之间的相互转换
- hd 1021 Fibonacci Again 规律对4取余为2 则yes
- 速配游戏