leetcode-problemset/leetcode-cn/originData/meChtZ.json

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        "question": {
            "questionId": "1000133",
            "questionFrontendId": "LCP 20",
            "categoryTitle": "Algorithms",
            "boundTopicId": 411214,
            "title": "快速公交",
            "titleSlug": "meChtZ",
            "content": "English description is not available for the problem. Please switch to Chinese.",
            "translatedTitle": "快速公交",
            "translatedContent": "小扣打算去秋日市集，由于游客较多，小扣的移动速度受到了人流影响：\r\n- 小扣从 `x` 号站点移动至 `x + 1` 号站点需要花费的时间为 `inc`；\r\n- 小扣从 `x` 号站点移动至 `x - 1` 号站点需要花费的时间为 `dec`。\r\n\r\n现有 `m` 辆公交车，编号为 `0` 到 `m-1`。小扣也可以通过搭乘编号为 `i` 的公交车，从 `x` 号站点移动至 `jump[i]*x` 号站点，耗时仅为 `cost[i]`。小扣可以搭乘任意编号的公交车且搭乘公交次数不限。\r\n\r\n假定小扣起始站点记作 `0`，秋日市集站点记作 `target`，请返回小扣抵达秋日市集最少需要花费多少时间。由于数字较大，最终答案需要对 1000000007 (1e9 + 7) 取模。\r\n\r\n注意：小扣可在移动过程中到达编号大于 `target` 的站点。\r\n\r\n**示例 1：**\r\n>输入：`target = 31, inc =  5, dec = 3, jump = [6], cost = [10]`\r\n>\r\n>输出：`33`\r\n>\r\n>解释：\r\n>小扣步行至 1 号站点，花费时间为 5；\r\n>小扣从 1 号站台搭乘 0 号公交至 6 * 1 = 6 站台，花费时间为 10；\r\n>小扣从 6 号站台步行至 5 号站台，花费时间为 3；\r\n>小扣从 5 号站台搭乘 0 号公交至 6 * 5 = 30 站台，花费时间为 10；\r\n>小扣从 30 号站台步行至 31 号站台，花费时间为 5；\r\n>最终小扣花费总时间为 33。\r\n\r\n\r\n**示例 2：**\r\n>输入：`target = 612, inc =  4, dec = 5, jump = [3,6,8,11,5,10,4], cost = [4,7,6,3,7,6,4]`\r\n>\r\n>输出：`26`\r\n>\r\n>解释：\r\n>小扣步行至 1 号站点，花费时间为 4；\r\n>小扣从 1 号站台搭乘 0 号公交至 3 * 1 = 3 站台，花费时间为 4；\r\n>小扣从 3 号站台搭乘 3 号公交至 11 * 3 = 33 站台，花费时间为 3；\r\n>小扣从 33 号站台步行至 34 站台，花费时间为 4；\r\n>小扣从 34 号站台搭乘 0 号公交至 3 * 34 = 102 站台，花费时间为 4；\r\n>小扣从 102 号站台搭乘 1 号公交至 6 * 102 = 612 站台，花费时间为 7； \r\n>最终小扣花费总时间为 26。\r\n\r\n\r\n**提示：**\r\n- `1 <= target <= 10^9`\r\n- `1 <= jump.length, cost.length <= 10`\r\n- `2 <= jump[i] <= 10^6`\r\n- `1 <= inc, dec, cost[i] <= 10^6`",
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