ACM大作业~~

我没退役啊,hahahhaa

写一些简答题

A

初等算术

Problem:B

Time Limit:1000ms

Memory Limit:10000K

Description

小学生在学多位数的加法时，是将两个数右对齐，然后从右往左一位一位地加。多位数的加法经常会有进位。如果对齐的位相加结果大于或等于十就给左边一位进一。对小学生来说，进位的判断是比较难的。你的任务是：给定两个加数，统计进位的次数，从而帮助老师评估加法的难度。

Input

输入文件中的每一行为两个无符号整数，少于10位。最后一行位两个0，表示输入结束。

Output

对输入文件(最后一行除外)每一行的两个加数，计算它们进行加法运算时进位的次数并输出。具体输出格式详见样例输出。

Sample Input

123 456555 555123 5940 0

Sample Output

No carry operation.3 carry operations.1 carry operation.

Hint

0时，输出No carry operation.   1时，输出1 carry operation.  大于1时，输出N carry operations.,注意：operations.和operation.

Source

#include <iostream>#include <string.h>using namespace std;int main(){    char s1[100],s2[100];    while(cin>>s1>>s2){        if(strcmp(s1,"0")==0&&strcmp(s2,"0")==0)            break;        int l1=strlen(s1);        int l2=strlen(s2);        int ans=0;        int i=l1-1,j=l2-1;        int tp=0;         int a;        for(;i>=0&&j>=0;i--,j--)        {            a=(s1[i]-'0')+(s2[j]-'0')+tp;            if(a>=10)            {                ans++;                tp=a%10;            }        }        if(i){            for(int k=i;k>=0;k--)            {                a=s1[k]-'0'+tp;                if(a>=10){                    ans++;                    tp=a%10;                }            }        }        if(j){            for(int k=j;k>=0;k--)            {                a=s2[k]-'0'+tp;                if(a>=10){                    ans++;                    tp=a%10;                }            }        }        if(ans==0)        {            cout<<"No carry operation."<<endl;        }        else if(ans>=2)        {            cout<<ans<<" carry operations."<<endl;        }        else if(ans==1){            cout<<ans<<" carry operation."<<endl;        }    }    return 0;}

老鼠的旅行

Problem:C

Time Limit:1000ms

Memory Limit:65536K

Description

一只老鼠有M磅猫食，然后在N个房间里面用猫食换JavaBean，房间i中能用F[i]磅的猫食来换J[i]磅的JavaBean，而且老鼠可以在一个房间里根据一定比例a%来换取JavaBean.现在他是这任务分配给你：告诉他，他的JavaBeans的获取能最多。

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1′s. All integers are not greater than 1000.M是开始时老鼠有的猫食！

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

5 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

Hint

贪心

Source

ZJCPC2004

#include <iostream>#include <algorithm>#include <stdio.h>using namespace std;struct node{    double f,j,v;} d[1005];bool cmp(node a1,node a2){    return a1.v>a2.v;}int main(){    double m;//f[1005],j[1005],;    int n;    while(cin>>m>>n)    {        if(m==-1&&n==-1)            break;        for(int i=0; i<n; i++)        {            cin>>d[i].f>>d[i].j;            d[i].v=d[i].f/d[i].j;        }        sort(d,d+n,cmp);        double ans=0.0;        for(int i=0; i<n; i++)        {           // cout<<d[i].v<<" "<<m<<endl;            if(m>=d[i].j)            {                m-=d[i].j;                ans+=d[i].f;            }            else            {                ans+=m*d[i].v;                m=0;            }        }        printf("%.3lf\n",ans);           }    return 0;}