发布时间 : 星期三 文章历年ACM更新完毕开始阅读
#include \
#define U 1000 #define M 101 //int c[M][3]={0}; struct bid {
char w[6];
int price;//价格 }b[M];
void sort(int m) {
int i,j;
struct bid bk; for(i=1;i for(j=i+1;j<=m;j++) { if(b[i].price>b[j].price) { bk=b[i]; b[i]=b[j]; b[j]=bk; } } } return; } int compute(int m) { int i,j,k,r,c[m][3]; j=1; c[1][0]=b[1].price;//记录 价格 c[1][1]=1;//记录价格出现的次数 c[1][2]=1;//记录是b数组中第几个人第一次出价 i=2; while(i<=m) { if(b[i].price==c[j][0]) { c[j][1]=c[j][1]+1; } else { j++; c[j][0]=b[i].price; c[j][1]=1; c[j][2]=i; } i++; } r=j; //按价格的出现次数为第一关键字，价格的高低为第二关键字排序 //次数由少到多，价格由低到高 for(i=1;i for(j=i+1;j<=r;j++) { if(c[i][1]>c[j][1]) { k=c[i][0]; c[i][0]=c[j][0]; c[j][0]=k; k=c[i][1]; c[i][1]=c[j][1]; c[j][1]=k; k=c[i][2]; c[i][2]=c[j][2]; c[j][2]=k; } } } for(k=1;k<=r;k++) { cout< return c[1][2]; } main() { int u,m,i,j,k; char w[6]; cin>>u>>m; for(i=1;i<=m;i++) { scanf(\ cin>>b[i].price; } sort(m); for(i=1;i if(strcmp(b[i].w,b[i+1].w)!=0) break; } if(i==m)//说明只有一个竞标者 { printf(\ printf(\ } else { k=compute(m); printf(\ printf(\ } } 【试题七】 BUYING FEED Farmer John needs to travel to town to pick up K (1 <= K <= 100)pounds of feed. Driving D miles with K pounds of feed in his truck costs D*K cents. The county feed lot has N (1 <= N <= 100) stores (conveniently numbered 1..N) that sell feed. Each store is located on a segment of the X axis whose length is E (1 <= E <= 350). Store i is at location X_i (0 < X_i < E) on the number line and can sell John as much as F_i (1 <= F_i <= 100) pounds of feed at a cost of C_i (1 <= C_i <= 1,000,000) cents per pound. Amazingly, a given point on the X axis might have more than one store. Farmer John starts at location 0 on this number line and can drive only in the positive direction, ultimately arriving at location E, with at least K pounds of feed. He can stop at any of the feed stores along the way and buy any amount of feed up to the the store's limit. What is the minimum amount Farmer John has to pay to buy and transport the K pounds of feed? Farmer John knows there is a solution. Consider a sample where Farmer John needs two pounds of feed from three stores (locations: 1, 3, and 4) on a number line whose range is 0..5: 0 1 2 3 4 5 --------------------------------- 1 1 1 Available pounds of feed 1 2 2 Cents per pound It is best for John to buy one pound of feed from both the second and third stores. He must pay two cents to buy each pound of feed for a total cost of 4. When John travels from 3 to 4 he is moving 1 unit of length and he has 1 pound of feed so he must pay 1*1 = 1 cents. When John travels from 4 to 5 he is moving one unit and he has 2 pounds of feed so he must pay 1*2 = 2 cents. The total cost is 4+1+2 = 7 cents. 【Standard input】 Line 1: Three space-separated integers: K, E, and N Lines 2…N+1: Line i+1 contains three space-separated integers: Xi Fi Ci 【Standard output】 A single integer that is the minimum cost for FJ to buy and transport the feed 【Sample input】 【Sample output】 2 5 3 7 3 1 2 4 1 2 1 1 1 #include \#include \#include \ #define N 101 struct store { int x; int f; int c; int cost; }s[N]; void sort(int n) { int i,j; struct store sk; for(i=1;i for(j=i+1;j<=n;j++) { if(s[i].cost>s[j].cost) { sk=s[i]; s[i]=s[j]; s[j]=sk; } } } return; }