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uva 11093

2015-09-06 00:46 357 查看
从第一个加油站开始遍历,如果到第k个加油站没有油了,那第1,2,3,4····k加油站肯定也不行 复杂度O(n)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int maxn = 200000+100;
int p[maxn], q[maxn];
int T, n;

int main()
{
int Case = 0;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
int res1 = 0, res2 = 0;
memset(p, 0, sizeof(p));
memset(q, 0, sizeof(q));
for(int i = 0; i < n; ++i) {
scanf("%d", &p[i]);
res1 += p[i];
}
for(int i = 0; i < n; ++i) {
scanf("%d", &q[i]);
res2 += q[i];
}
printf("Case %d: ", ++Case);
if(res1 < res2) { printf("Not possible\n"); continue; }
int sum = 0, id = 1, is = 0;
for(int i = 0; i < n; ++i) {
sum += p[i];
sum -= q[i];
if(sum < 0) {
sum = 0;
id = i+2;
}
}

printf("Possible from station %d\n", id);
}
return 0;
}
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