usaco1.1.*
来源:互联网 发布:linux重启后进不去系统 编辑:程序博客网 时间:2024/06/15 01:05
usaco1.1.1
char a[7]; char b[7]; FILE *fin =fopen("ride.in", "r"); FILE *fout =fopen("ride.out", "w"); long i, ca = 1, cb =1; fscanf(fin , "%s %s", a,b);
for(i = 0; a[i];i++) ca *= a[i] - 64; for(i = 0; b[i];i++) cb *= b[i] - 64; if(ca % 47 == cb %47) fprintf(fout, "GO\n"); else fprintf(fout, "STAY\n");
return 0;
char name[15]; int count; int num; int money; int people; int k, p; char name2[15]; gift gift1[15]; FILE *fin =fopen("gift1.in", "r"); FILE *fout =fopen("gift1.out", "w"); fscanf(fin, "%d",&num); for(int i = 0; i< num; i++) { fscanf(fin, "%s", gift1[i].name); gift1[i].count = 0; } for(int i = 0; i< num; i++) { fscanf(fin, "%s", name2); fscanf(fin, "%d%d", &money,&people); if(people == 0) continue;
for(p = 0; p < num; p++) { if(!strcmp(gift1[p].name, name2)) break; } gift1[p].count -= (money - money%people);
for(int j = 0; j < people;j++) { fscanf(fin, "%s", name2); for(k = 0;k < num; k++) { if(!strcmp(gift1[k].name,name2)) break; } gift1[k].count += money/people; } } for(int i = 0; i< num; i++) { fprintf(fout, "%s %d\n", gift1[i].name,gift1[i].count); } return 0;
int n; int year; int t = 13; int z; int b[8]; int a[15] = {31, 31, 28,31, 30, 31, 30, 31, 31, 30, 31, 30}; FILE *fin =fopen("friday.in", "r"); FILE *fout =fopen("friday.out", "w"); fscanf(fin, "%d",&n); memset(b, 0,sizeof(b)); for(int i = 0; i< n; i++) { year = 1900 + i; if((year % 4 == 0&& year % 100 != 0) || (year % 100== 0 && year % 400 == 0)) a[2] =29; else a[2] = 28; if(i == 0) a[0] =0; else a[0] = 31; for(int j = 0; j < 12; j++) { t +=a[j]; z = t %7; b[z]++; } } fprintf(fout, "%d %d %d%d %d %d %d\n", b[6], b[0], b[1], b[2], b[3], b[4], b[5]); return 0;
int n,a=0,b=0,w=0,m=0; char s[701]={0}; FILE*in=fopen("beads.in","r"); FILE*out=fopen("beads.out","w"); fscanf(in,"%d",&n); fscanf(in,"%s",s); int i,j,k; char now=0; memcpy(s+n,s,n); for(i=0;s[i]&&m<n;i++) { if(s[i]=='w') { w++; b++; } else if(now!=s[i]) { if(a+b>m) m=a+b; a=b-w; b=w+1; w=0; now=s[i]; } else { b++; w=0; } } if(a+b>m) m=a+b; fprintf(out,"%d\n",m>n?n:m); return0;
#include <stdio.h>
int main()
{
}
usaco1.1.2
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
typedef struct
{
}gift;
int main()
{
}
usaco1.1.3
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
int main()
{
}
usaco1.1.4
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int main()
{
}
- usaco1.1.*
- USACO1.4.4
- USACO1.4.2
- USACO1.3.1
- usaco1.1
- usaco1.2.*
- USACO1.1.4
- USACO1.5.4 checker challenge
- USACO1.2.2 transfomations
- USACO1.5.4 checker challenge
- usaco1.1 beads
- USACO1.2 Milking Cows
- USACO1.1.4 Broken Necklace
- USACO1.2.1 Milking Cows
- USACO1.2.2 - transform
- usaco1.2.6 Dual Palindromes
- usaco1.1.2的题解
- usaco1.1.3的题解
- Qt精彩实例(4)& 自己实现的软件管家界面
- 医疗产业走进大数据时代:祸兮?福兮?
- 栈,链表
- 带外数据
- KMP
- usaco1.1.*
- 认识六个被误解的 Ruby 特性
- usaco1.2.*
- 二叉树遍历
- 链表插入,删除,排序,反转
- 拓扑排序, 快速排序, 冒泡排序, 堆排序, 二分查找
- 2012年12月30日
- HDU 1074 Doing Homework(状态DP)
- 正则表达式